What is the work done when pressure fully changes in thermodynamics?

Under very small changes in $V$, the pressure can be taken as approximately constant over that interval, so it is possible that this equation will sometimes be used taking $p$ constant.

However, in the general case you are 100% right that $p$ is not constant. The trick, then, is knowing how $p$ depends on $V$ so you can calculate $W$ with the integral $$W = -int p dV.$$

An especially good example of this is for an ideal gas, where you know that $PV = NkT$, so $P = NkT/V$. There are two (simple) cases here: isothermal expansion, where $T$ is held constant as $V$ increases, and adiabatic expansion, where no heat is allowed to enter the system. I suggest looking into how the work is derived in these situations; this is certainly how I came to understand this slightly subtle issue.