FBD for Cylinder in water , acceleration of water and why is $P$ perpendicular force? 
Here, we have accounted two kinds of pressures according to my book.
One is atmospheric pressure and other is pressure exerted by liquid of density $f$.
My questions are that
first we say that pressure is a perpendicular force and so force will be upwards from surface $PQ$. I want to know is it not downwards since $g$ is going to always pull it down?
Second, if we take consideration into Newton’s third law, is it that by total force on the surface $PQ$, we can find acceleration of water?
Third, How should we draw its FBD?

 A: 
first we say that pressure is a perpendicular force and so force will be upwards from surface PQ.I want to know is it not downwards since g is going to always pull it down.

Graity is pulling down on the water above the bottom surface and the atmosphere is also pushing down on the surface of the water, resulting in
$$\tag 1 P_{PQ} = P_{RS} + \rho g h$$
where $g$ is the acceleration due to gravity and does indeed act down (to the centre of the earth). The pressure at the bottom of the surface $P_{PQ}$, is due to the pressure of the atmosphere $P_{RS}$ plus the pressure due to the weight of the fluid.

Second, if we take consideration into Newton’s third law , is it that by total force on the surface PQ , we can find acceleration of water.

The water is not accelerating since the fluid is static. The weight of the fluid is equal to $mg$. Assuming the density remains constant, the weight can be calculated using the density $\rho$ and the area $A$ so that
$$W = mg = \rho Vg = \rho Ahg$$
where $A$ is the area of the circle $PQ$ (or $RS$).

Third , How should we draw its FBD.

You can do this by noting the following.
We can write equation (1)
$$p_{PQ} = p_{RS} + \frac{\rho Ahg}{A}$$
and in terms of forces
$$F_{PQ}= F_{RS}  + \rho Ahg$$
or
$$\rho Ahg = F_{PQ}- F_{RS}$$
which is the same as
$$W = F_{PQ}- F_{RS}$$
It makes sense that the weight of the column is equal to the force at the bottom minus the force at the top since the column is in static equilibrium.
