Solution:

We make use of the fact that an angle inscribed in a circle has measure equal to one-half of the arc subtended. Since the $x$- and $y$-axes meet in a right angle, the circle $C_{1}$ through $B$, $O$, and $Q$ has $Q B$ as a diameter. Also, $\angle A P B$ is a right angle, since $A B$ is the diameter of $C$. But this means that $\angle Q P B$ and $\angle Q O B$ are both right angles, so that $P$, $O$, $Q$, $B$ all lie on circle $C_{1}$. Thus the two angles in question, $\angle B Q P$ and $\angle B O P$, are inscribed in $C_{1}$, subtend the same arc, and are therefore equal.