Consider a fluid filled in a cylindrical container, height $h$ and area of cross-section $a$. The pressure at the bottom will be:
$$p_0+\rho gh$$
but not
$$p_0+\rho gh+\frac{mg}{a}$$ where $m$ is the mass of fluid in the container and $\rho$ being its density. Why is this so?
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.
=>$$\rho gh=\frac{mgh}{v}$$ =>$$\rho gh=\frac{mg}{a}\ \ (since\ h\times a=v)$$
As you can see you counted the term twice.
