This is the same principle as the reduction of NAD+ to NADH, which is especially MCAT-relevant. The most common way this reaction is written out is:
NAD+ + 2e- + H+ ----> NADH
As you can see, NAD+ is gaining electrons to become its reduced form, NADH. But it can be confusing to think about where these electrons come from, since H+ itself has one proton and zero electrons. But remember, hydrogen isn't always in the form of H+. It can take the form of hydride (H-) and attack NAD+. Since hydrogen has one proton, a hydride anion must have one proton and two electrons. NAD+ gains these two electrons as it becomes NADH. Alternatively, you can think about NAD+ reacting with H2. H2 has two protons and two electrons in total; NAD+ keeps one of the protons and both the electrons, and releases the remaining H+. Again, this same principle works for the reduction of NADP+ as well as NAD+.
Going back to the original post here - remember, these are redox reactions. If NADP+ were to gain an H+ and become NADPH2+ (which, again, is not what happens), that would not be a redox reaction.
Good luck with your MCAT prep!