All of the other answers are correct. However I shall re-express their answer in terms of "weight of fluid above you", which perhaps you will find more palatable.
For simplicity say the solid body submerged in the fluid is an upright cylinder of cross-sectional area $A$, and has weight $W_{cylinder}$. We want to calculate pressure at some point in the fluid beneath the cylinder. To do so we shall calculate weight of fluid+solid-cylinder column above that point, acting on horizontal area of magnitude $A$, and then divide total force acting on that area by magnitude of area. Key to the argument is that total force acting on area is not simply equal to weight of fluid +solid-cylinder column above that area.
Total weight of fluid+solid-cylinder column=$W_{cylinder}+W_{fluid}$
But since the cylinder is held stationary, then by virtue of Archimedes principle, we must exert an upward force on it equal to $W_{cylinder}-W_{displaced~fluid}$, where $W_{displaced~fluid}$ is simply the weight of fluid displaced by cylinder.
So net downward force on area $A$ is \begin{align} (W_{cylinder}+W_{fluid})-(W_{cylinder}-W_{displaced~fluid})=W_{fluid}+W_{displaced~fluid} \end{align} which is nothing but weight of water column in which the solid cylinder has been replaced by fluid cylinder. That is why, to calculate pressure, it does not matter whether there is a solid object hanging above a point inside the fluid, provided the whole system is in equilibrium.
P.S. I have assumed that cylinder is denser than fluid. I shall leave it to you to generalize the argument.