You should also notice that since $CP$ is the diameter of the circle, angle $CBP$ is a right angle.
says the solution to this question. But what is the reasoning or proof behind the claim that the angle opposite the diameter ($\angle CBP$) must be a right angle?
I started to draw it out to gain some intuition, but I only got about as far showing that drawing a line from the center $O$ to the vertex of $CBP$ splits the inscribed triangle into two, where $\angle BOC + \angle BOP = \pi$. I'm not sure if that will lead to a proof, or whether this is more easily shown with e.g. the Law of Sines.