Pressure ($P$) is the force ($F$) per unit area ($A$): $$ \begin{equation}\begin{aligned} P=\frac{F}{A}\\ \end{aligned}\end{equation} $$
The SI unit is: $$ \begin{equation}\begin{aligned} P&=\frac{F}{A}\\ &\rightarrow \frac{N}{m^2}\\ &=Nm^{-2}\\ \end{aligned}\end{equation} $$
Example: Find the pressure exerted by a $5kg$ block placed on a surface. The area of the block in contact with the surface is $2m^2$.
Solution: $$ \begin{equation}\begin{aligned} P&=\frac{F}{A}\\ &=\frac{5kg\times 9.81 ms^{-2}}{2m^2}\\ &=24.5Nm^{-2}\\ \end{aligned}\end{equation} $$
Fluid pressure
Given a fluid of density $\rho$ and an object submerged at a depth of $h$, the pressure exerted by the fluid on the object is: $$ \begin{equation}\begin{aligned} P=\rho gh\\ \end{aligned}\end{equation} $$
If the object moves from a depth of $h_1$ to a new depth of $h_2$, the pressure change is: $$ \begin{equation}\begin{aligned} \Delta P&=\rho g(h_2-h_1)\\ &=\rho g\Delta h\\ \end{aligned}\end{equation} $$
Total pressure for an open vessel
If we have an object that is submerged in a fluid contained in an open vessel (a container exposed to the atmosphere), then the total pressure on the object is the sum of the fluid pressure and atmospheric pressure: $$ \begin{equation}\begin{aligned} P_{total}&=P_{fluid}+P_{atmos}\\ &=\rho gh + P_{atmos}\\ \color{royalblue}P_{total}&=\rho g\color{green}h \color{normal}+ P_{atmos}\\ \color{royalblue}y&=m\color{green}x\color{normal}+c\\ \end{aligned}\end{equation} $$
This means that if we plot a graph of $P_{total}$ versus the depth $h$ of the object, the gradient of this graph will be $\rho g$ and the y-intercept will be $P_{atmos}$.