Consider an element of fluid at rest within a rectangular co-ordinate system defined by three axes, $x$, $y$, and $z$. The element is at rest (not moving), so there must be no overall (net) force acting on it - the various forces must be in balance. These forces include that arising from pressure acting downward on the top face of element, that arising from pressure acting upward on the bottom face of the element, and the (downward) force due to gravity acting on the mass of the element. Just to clarify the nomenclature, $F_{z,z}$ means the force in the $z$-direction at the point $z$; $F_{z,z+\delta z}$ means the force in the $z$-direction at the point $z+\delta\,z$.
The force in the $z$-direction at the point $z$ is given by multiplying the pressure at that point, $P_z$, by the area over which the pressure operates $A$,
$$F_{z,z}=P_zA\qquad\text{-upward on the bottom face}$$
Similarly, the force in the $z$-direction at the point $z+\delta z$ is given by,
$$F_{z,z+\delta z}=P_{z+\delta z} A\qquad\text{-downward on the top face}$$
The downward force due to gravity is given by (mass) $\times$ (acceleration due to gravity), where (mass) = (density $\times$ volume),
$$F_g=mg=\rho g\delta z A$$
A force balance then gives,
$$F_{z,z+\delta z}+F_g=F_{z,z}$$
$$P_{z+\delta z}A +\rho g\delta z A=P_z A$$
Rearranging and dividing through by volume $\left(\delta z A\right)$,
$$\frac{P_{z+\delta z}-P_z}{\delta z}=-\rho g$$
Setting the limit on the left hand side as $\delta z \rightarrow 0$ gives the definition of the derivative of $P$ with respect to $z$,
$$\frac{\text{d}\,P}{\text{d}\,z}=-\rho g$$
In other words, the change in pressure, $P$, with height, $z$, is negative - pressure reduces with height (or increases with depth).
Then, to find the pressure at a particular position relative to the pressure at the surface, we integrate from the surface,
$$\int^P_{P_0}\text{d}\,P=-\int^z_0\rho g\text{d}\,z$$
If the density of the fluid is constant with $z$-position, and if $g$ does not change with height (which it does not, over most changes in height of practical
interest), then,
$$P-P_0=-\rho gz$$
Note that $z$ is negative as we go downwards - let's replace this with $h$, depth, which is positive as we go downwards, i.e. $h = -z$,
$$P-P_0=\rho gh$$
$$P=P_0+\rho gh$$
The interactive graph below looks at this equation with 3 fluids where the height and density of each can be changed.