Fluid statics and and pressure

We know that, at a depth $h$ below the surface of water, pressure will be exerted by the atmosphere and the weight of water present till that level $h$, so pressure should be $P_h = (P_o + frac{mg}{a})$

BUT: $rho g h =frac{mg}{a}$

Proof:

$m = rho V$, where $rho$ is density, $V$ is volume

We know that, $V = ah implies m = rho ah$

Therefore, $frac{mg}{a} = frac{(rho a h) g}{a} = rho g h$

This proves that $boxed{P_h = P_o + rho g h}$

This is because $rho gh$ is itself the pressure due to the liquid on the bottom of the container.

=>$$rho gh=frac{mgh}{v}$$ =>$$rho gh=frac{mg}{a} (since htimes a=v)$$

As you can see you counted the term twice.