Pressure of the Fluid Depending on the Depth when an Object is Placed on the Fluid
Explanation
Simply put, when an object is on a fluid, the pressure depending on the depth can be obtained by substituting ${P_{0} }^\prime$ for $P_{0}$ in the case of calculating the pressure in a fluid depending on the depth. The original formula’s atmospheric pressure $P_{0}$ represented the pressure exerted from above the fluid. That is, if an object is placed on the fluid, adding the pressure due to the object to the atmospheric pressure gives the pressure exerted above the fluid. It would be easier to understand if you think of it as the atmospheric pressure increasing without considering the object. Now, let’s calculate ${P_{0}}^\prime$. The object on the fluid is in equilibrium, so summing all the forces acting on it gives $0$. Let’s say the mass of the object is $M$, the area of the object’s top/bottom surface is $A$, and the direction the same as gravity is $+$.
Gravity$\downarrow $ + Force due to pressure on the top surface (force due to atmospheric pressure)$\downarrow$+ Force due to pressure on the bottom surface (force due to pressure at the fluid surface) $\uparrow $ $=0$ $$ \implies Mg+P_{0}A-{P_{0}}^\prime A=0 $$
$$ \implies Mg/A + P_{0} – {P_{0}}^\prime = 0 $$
$$ \implies {P_{0}}^\prime = P_{0} + \dfrac{Mg}{A} $$
Therefore,
$$ P_{h}={P_{0}}^\prime + \rho g h = P_{0} + \rho gh + \dfrac{Mg}{A} $$