Why voltage is not the same for the capacitors in series? 
In this picture, there are two capacitors C1 and C2 joined in series and connected to a battery. We know there are two terminals in a battery, a positive terminal and a negative terminal.
The potentials of the positive and negative terminals are +P and -P respectively. And there potential difference, in other word the voltage of the battery is V.
The plate on the left side of the capacitor C1 is directly connected to the positive terminal of the battery. So its potential will also be +P. Similarly the potential of the plate on the right side of C2 will have a -P potential.
My questions:
A) I am assuming the magnitudes of the potentials of the opposite sides of the battery are same just different in signs. For the right plate of C2 to have a potential -P it takes -q amount of charge. The left plate of C1 also gives off this equal amount of charge and acquires the state +q. Since C1 and C2 are two different capacitors, why gain or loss of same amount of charge would cause them to have the same magnitude of potential?
B) If the left plate of C1 has a charge of +q and a potential +P, then C1's other plate should also have the potential -P as it has -q charge. Because two plates of the capacitor C1 are same in material and geometry. Similarly left plate of C2 has to acquire to the potential +P. If this happens then the voltage of C1 and C2 will be same V and net voltage of the two capacitors will become 2V which is wrong.
If someone could clarify this confusion it would be a big help.
 A: 
A) ....Since C1 and C2 are two different capacitors, why gain or loss of same
amount of charge would cause them to have the same magnitude of
potential?

They have the same amount of charge but the magnitudes of the voltages (potential differences) of the capacitors cannot be the same if the capacitances are not the same. The relationship between voltage, capacitance and charge for a capacitor is given by the equation
$$C=\frac{Q}{V}$$
Where $Q$ is the charge on either plate ($+q$ or $-q$ in your circuit).
therefore
$$Q_{1}=C_{1}V_{1}$$
$$Q_{2}=C_{2}V_{2}$$
For series capacitors
$$Q_{1}=Q_{2}$$
therefore
$$V_{1}C_{1}=V_{2}C_{2}$$
$$V_{1}=\frac{C_2}{C_1}V_{2}$$

B) If the left plate of C1 has a charge of +q and a potential +P, then
C1's other plate should also have the potential -P as it has -q
charge. Because two plates of the capacitor C1 are same in material
and geometry.

That is not correct. Assuming an ideal wire (zero resistance) connecting the capacitors, the potential on the right plate of $C_1$ has to be the same as the potential on the left plate of $C_2$ because electrically they are the same point. The same point can't have different potentials.
If the potentials were $-P$ and $+P$ as you say, then if you apply Kirchhoff's voltage law (KVL) you will find, as follows, that $P$ would have to be zero.
Label the positive and negative terminals of the battery $a$ and $b$. Then the potential difference $V$ of the battery is
$$V=V_{ab}=V_{a}-V_{b}=P-(-P)= 2P$$
Now apply KVL clockwise beginning with the battery:
$+V-V_{C1}-V_{C2}=0$
$+2P-[P-(-P)]-[+P-(-P)]=0$
$-2P=0$
$P=0$

How can the right of C1 and left of C2 have the same potential? I get
that they are connected with a wire and must have the same potential;
but they also have the opposite charge, I just cannot understand this
idea.

The problem is you are looking at the charge on the right plate of C1 and the left plate of C2 in isolation.
Each capacitor has an equal amount of positive and negative electric charge so each capacitor is electrically neutral. That means the electric field between the capacitors has to be zero. Since the electric field between two points is the gradient (rate of change) of potential between the points, the potential difference between the right plate of C1 and left plate of C2 has to be zero.
Hope this helps.
A: The sign we label on both plates of a capacitor represents the sign of charge accumulated on that plate, which is not an indicator of the sign of electric potential. In fact, the electric potential can take any values (the potential across the battery can be from $V_0$ to $0$, or from $0$ to $-V_0$, or even from $100V_0$ to $99V_0$ if you like) as long as you ensure the potential difference between points of interest is correct. I'll stick with your notation of potential from $+p$ to $-p$ across the battery, though.
In equilibrium, the potential on the left of $C_1$ is $+p$, and that on the right of $C_2$ is $-p$. However, the potential in the middle (from the right plate of $C_1$ to the left plate of $C_2$) should be a value in between, call in $V_a$. The value of $V_a$ should make sure that the charge accumulates on both capacitors are the same (conservation of charge), which requires $C_1(p-V_a)=C_2(V_a-(-p))$. In case of different $C_1$ and $C_2$
, this leads to different potential differences across the two capacitors ($p-V_a$ and $V_a+p$).
For Question 1: True, they have the same amount of charge accumulation, but they have different potential differences.
For Question 2: The statement "then C1's other plate should also have the potential -P as it has -q charge. Because two plates of the capacitor C1 are same in material and geometry. Similarly left plate of C2 has to acquire to the potential +P." is false. If this is really the case, then the right plate of $C_1$ and the left plate of $C_2$ are in different potentials. But this is impossible since they are connected by a wire and in equilibrium. To really answer this question, the symmetry between two plates of a capacitor does not dictate that it should have potentials in opposite signs but with equal magnitude. As I mentioned before, the electric potential is just a relative measure and you can shift the coordinate in whatever ways you like. For example, you can shift the coordinate so that the left and the right plate of $C_1$ are in potentials that are "opposite in sign with equal magnitude". Doing so makes the right plate of $C_2$ have a potential that is more negative. After all, you just need to ensure that the potential difference between the left plate of $C_1$ and the right plate of $C_2$ is the voltage across the battery, as required by Kirchoff's loop rule.
A: This is an important point which is commonly overlooked. The crux is that the voltage drop on each capacitor is not necessarily the batteries voltage. Let's work out the ideas step by step.
Firstly, to find the charge on capacitor system let us take equivalent capacitance. When we replace the two with an equivalent one, all voltage will drop on that, hence the voltage of that capacitor will be same as the battery.
Finding equivalent capacitance:
$$ C_{eq} =  \frac{C_1  C_2}{C_1 + C_2}$$
Now, we can find the charge, which is basically by saying whole voltage is dropped on the equivalent capacitor:
$$ V \frac{C_1C_2}{C_1 + C_2} = Q$$
Note that charge on equivalent capacitor would be same on each individual capacitor which made it, let us find voltage on each capacitor starting with $C_1$ via the equation $ V = \frac{Q}{C}$
$$ V_1 =\frac{ \frac{VC_1 C_2}{C_1 + C_2}}{C_1} = \frac{VC_2}{C_1 + C_2}$$
Similarly,
$$ V_2 = \frac{VC_1}{C_1 + C_2}$$
We find that each doesn't drop the same amount of voltage, and after charging the charge built up is such that $V_1 +V_2=V$.
P.S: This idea also apply in series resistor, suppose two resistor in series connected to battery. Try applying the 'incorrect logic' to that, you will find the same error.
