Here is the diagram for the problem and the question is asking me to find the angular velocity of the rod when it is completely vertical. I used the conservation of energy but I was wondering why I have to choose (l/2) as the maximum height for PE and not use $MgL$. What I was thinking is that since it is $L$ higher than its initial position that means the height is $L$.
It is a shortcut that is usually derived once in introductory physics classes, but as usual the derivation is forgotten but the concept sticks.
If you have your rod in a uniform, vertical field, each mass element $\text dm$ has a potential energy of $\text dU=gy\,\text dm$, where $g$ is the acceleration due to the field and $y$ is the vertical height of the mass element (where, as usual, where $y=0$ can be set anywhere). Then, for a vertical rod of uniform density $\lambda=\text dm/\text dy=M/L$ whose base is at $y=y_0$, the total potential energy of the rod is $$U_\text{vertical}=\int\text dU=\int_{y_0}^{y_0+L}\lambda gy\,\text dy= \frac12\lambda g\left[(y_0+L)^2-y_0^2\right]=\frac12MgL+Mgy_0$$
For the horizontal rod located at $y=y_0+L$, the potential energy is a lot easier to determine since all parts of the rod are at the same height; it is just $U_\text{horizontal}=Mg(y_0+L)$ (you could set up another integral for this if you wanted to). Therefore, the change in potential energy from horizontal to vertical is $$\Delta U=U_\text{vertical}-U_\text{horizontal}=\frac12MgL+Mgy_0-Mg(y_0+L)=-Mg\frac L2$$
which is what is used in the problem.
