Rotational analogue of Newton's 2nd law If we are writing torque equation about general point what points must be taken into account for getting the correct result.
Also If that general point is accelerated then how do we deal with this.
Like the first one that should be taken into account is to have moment of inertia about that general point.
Like in given below problem: 
We have to calculate initial angular acceleration for above cases :


*

*if friction is absent.

*If friction is absent and pure rolling.

*If coefficient of friction is not enough to provide pure rolling.


For example how can we apply torque equation about point "P" to find initial angular acceleration.
Figure given in image below:
If i understood this question then my concepts would be all clear. 
I would really appreciate your help.
I'm adding original question and it's solution. I found it confusing as they used point C and didn't mentioned any pseudo force. Do you guys approve this.

.


 A: $\def\rA{{\rm A}} \def\rB{{\rm B}} \def\rP{{\rm P}} \def\br{{\bf r}} 
\def\bv{{\bf v}} \def\bF{{\bf F}} \def\bL{{\bf L}} \def\bM{{\bf M}} \def\D#1#2{{d#1 \over d#2}}$
I've been in SE for less than six months but I've seen that questions
about rotatory motions, torques, moments of inertia and so on are among
the most frequent (perhaps only beaten by relativity questions). I
understand the matter is not easy but I have also some doubts about how
it's taught. I'm afraid there are some delicate points aren't often
correctly discussed.
My first post on this subject may be found [here]
(For a solid sphere rolling (pure roll) up a slope (with friction) does friction play a role in slowing it down?). In this answer I'll not give proofs or an accurate discussion. I prefer to focus on the most relevant points and give what I think is the right approach.

First principles
1) I'm not going to use the term "torque" - it may be cause of
confusion. I prefer to speak of moment of a force.
2) In most problems there's no use shifting to a different frame. I'll always remain in one and the same (inertial) frame. So no pseudo-forces are present.
3) A moment (of forces, of momentum, or else) always requires a
reference point, not to be confused with a reference frame. In my
native language (italian) there is a term ("polo") to name that point
but I can't use its english translation (pole) as it has other meanings too. I'll use the shorthand RP.
4) If A is the RP a force $\bF$ applied in point P has a moment
$$\bM_\rA = (\br_\rP - \br_\rA)  \times \bF$$
It's independent of possible A's motion, velocity, acceleration, if
not because a change in A's position will change $\bM_\rA$.
5) If B is another RP, $\bM_\rB$ is defined analogously and will
generally differ from $\bM_\rA$. That's all.
6) The same also holds for angular momentum (i.e. moment of momentum).
For a mass point P it's defined as
$$\bL_\rA = (\br_\rP - \br_\rA) \times (m\,\bv_\rP).$$
For a different RP, say B, you have only to change $\br_\rA$ into $\br_\rB$.
It hasn't the least relevance if A and B have different motions:
their velocities don't appear in definitions. Of course if A (or B)
are moving $\bL_\rA$ will change in time not only because of P's
motion but also of A's. This is automatically taken into account in the definition.
7) If you aren't  given a single point but a system both $\bM$ and
$\bL$ are to be summed on all points. In computing $\bM$ only external
forces need to be taken into account.

The main equation (torque equation)
And now the most important point. When does equation
$$\D\bL t = \bM \tag1$$
hold true? Is it true for any RP? 
The answer is no. It holds


*

*if RP is the com of system, whatever its motion

*if RP velocity is parallel to the one of com: in particular if it's
the same or if it's null (still RP).
Don't worry about an acceleration of RP, but remember: you must stick
to one (inertial) frame, and all velocities must be computed in that
frame, even if the RP is moving.

The half-disk
Let's apply all this to your half-disk. You may use eq. (1) taking as RP


*

*the point P

*the com of half-disk (is it X? you didn't say).
It would be wrong to use C as it's moving with a velocity not parallel
to com's.
As to P, there's no problem for its being - so to say - split in two
after the half-disk rotates. You must keep clear that in computing
moments only one instant of time matters, not what happens afterwards.
Note: I'm assuming you know that if $\bL$ is computed by $I\omega$
then the right $I$ must be taken ($\omega$ stays the same irrespective
of your choice of RP).

Comments
Of course I'm worried about what you write:

I got the correct answer via applying from centre of the circle of
  full disc but I'm not getting via lower point.

This is in contradiction with what I wrote above. Why do you say so?
Did you compare your solution with another, maybe your teacher's?
Unfortunately you didn't tell us what question you were requested to
answer. I may only guess: "If the half-disk is left alone at rest in
the shown position, it will begin to fall. What coefficient of friction
is required in order that it doesn't slip on the ground?"
My answer is $\mu\ge16/(15\pi)$. What was yours?
Edit
There was an error: $\mu\ge8/(9\pi)$.
A: 
We first write the EOM's for the center  mass $C_m$
Translation 
$$m\,\vec{a}=\vec{F}\tag 1$$
and
Rotation 
$$I_{cm}\,\vec{\alpha}_{cm}+\vec{\omega}_{cm} \times (I_{cm}\,\vec{\omega}_{cm})=\vec{\tau}\tag 2$$
Where :
$\vec{a}$ translation acceleration
$\vec{\alpha}_{cm}$ angular acceleration
$\vec{\omega}_{cm}$ angular velocity
$\vec{F}$ sum of all apply forces
$\vec{\tau}$ sum of all apply torque
$I_{cm}$ Inertia Tensor in center of mass coordinate system
But we can write equation (2) for point $P$
$$I_{p}\,\vec{\alpha}_{p}+\vec{\omega}_p \times (I_{p}\,\vec{\omega}_p)=\vec{r}\times \vec{F}+\vec{\tau}\tag 3$$
with : $I_{p}$ Inertia Tensor in $p$ coordinate system
$I_p=I_{cm}+m\,\tilde{\vec r}^2$
and 
$\tilde{\vec r}=\left[ \begin {array}{ccc} 0&-r_z&r_y\\ r_z&0&-r_x
\\ -r_y&r_x&0\end {array} \right] 
$
A: The torque equation
$$
\text{net torque = change of net angular momentum per unit time}
$$
can be used with any fixed point of space as reference point, in inertial reference frame. 
In case 1, where there is no friction, the center of mass will move along straight line downwards, while the lowest mass point of the body moves to the left. Thus the center of rotation is initially at C. We can choose this point of space as our fixed point of reference. At $t=0$, there is only single force that has torque around this point, the gravity force.
In case 2, the body is rolling, so at $t=0$, the lowest point of the body in contact with the ground must not move with respect to the ground. Thus the center of rotation is this contact point and it is easiest to use this point of space as our fixed point of reference.
In case 3, there is some friction force due to the ground, which points to the right horizontally. Actual center of rotation depends on how strong this force is, so we cannot point beforehand where this point is. Then we have to choose a different point of reference. The most advantageous point seems to be the point that initially coincides with the center of mass: at $t=0$, torque of gravity is zero there.
A: $\let\om=\omega \def\rC{{\rm C}} \def\rP{{\rm P}} \def\rX{{\rm X}} \def\ns#1#2{#1_{\rm{#2}}} \def\Pd{\ns Pd} \def\Pg{\ns Pg} \def\dx{\dot x} \def\dy{\dot y} \def\ddx{\ddot x} \def\ddy{\ddot y} \def\bL{{\bf L}} \def\bM{{\bf M} \def\half{{\textstyle {1 \over 2}}}} \def\IPg{\ns I{P_g}} 
\def\D#1#2{{d#1 \over d#2}}$

Can you solve first case taking P as RP

So at last you let us know what questions were! I'll only study $\mu=0$,
which isn't so easy. To understand the answer it's best not to consider $t=0$ alone, but a generic $t$ before.
To take P as RP requires qualification. Because of slipping P must be
viewed as two different points: one P$_\rm g$ being the initial contact
point as seen belonging to ground, the other P$\rm d$ belonging to disk.
Only the former is fixed and qualifies as RP.
Let me introduce some coordinates: origin in P$_\rm g$, $x$-axis rightwards, $y$ axis upwards. Point C is constrained moving horizontally, so $\rC=(x,r)$ with $x(0)=0$, $\dx(0)=0$. Since no external horizontal forces are acting on half-disk, and X's initial velocity is null, its $x$ coordinate is fixed at $p=4r/(3\pi)$. Then $\rX=(p,y)$ with $y(0)=r$, $\dy(0)=0$.
In order to write $d\bL/dt=\bM$ with P$_\rm g$ as RP we only need to
compute $\ns I{P_g}$:
$$\ns I{P_g} = \ns IX + m\,\overline{\rm P_gX}^2 = 
\half m\,r^2 + m\,(p^2 + r^2).$$
Note that $\IPg$ is a function of $t$ because of $y$. But
$$\D{}t \IPg = 2 m\,y\,\dy$$
which vanishes at $t=0$. 
We're almost ready. But an important simplification may be made
because the half-disk's motion is of a special kind: a plane motion.
By this is meant that all velocities are parallel to one plane (the
$(x,y)$ plane in our case) so that angular velocity and acceleration
as well as angular momentum are orthogonal to that plane, i.e. are
directed along $z$-axis. The same holds for force moments. Then there
is no need to deal with vectors: $\om$, $M$, $L$ are simple scalars.
Moreover only axes parallel to $z$ are of interest in computing
moments of inertia, and the relationship between angular velocity and
angular momentum is simply written
$$L = I\,\om.$$
The only external force having not vanishing moment wrt P$_\rm g$ is
weight, so that
$$M = -m\,g\,x.$$
As to angular momentum
$$L = \IPg\,\om$$
$$\D Lt = \D{}t \IPg\,\om + \IPg\,\D \om t.\tag1$$
At $t=0$ we have $x=p$, $y=r$ whereas first term in right-hand side of (1)
vanishes. Then
$$\D \om t = {m\,g\,p \over \IPg} = 
    {2\,g\,p \over 3\,r^2 + 2\,p^2}.$$
Caution: I can't ensure no errors are present. Please check my calculations!
