Very nice problem! Let $A_{1}$ and $B_{1}$ be the second points of intersection of circumcircles of $\triangle AQP$ and $\triangle PQB$ with $AC$ and $BC$ respectively. $CP \cdot CQ = CB \cdot CB_{1}$ and $CP \cdot CQ = CA \cdot CA_{1} \Rightarrow CB \cdot CB_{1} = CA \cdot CA_{1}$. Since $CA = CB$ we got $CB_{1} = CA_{1}$ and then $\frac{CB}{CB_{1}} = \frac{CA}{CA_{1}} \Rightarrow A_{1}B_{1} \| AB$. In cyclic quadrilateral $AA_{1}QP$ we got $\angle AQP = \angle AA_{1}P$ but $ \angle AQP = \angle ACB \Rightarrow \angle AA_{1}P = \angle ACB \Rightarrow A_{1}P \| BC \Rightarrow \frac{AA_{1}}{A_{1}C} = \frac{AP}{PB} = 2$. So $\frac{BB_{1}}{B_{1}C} = \frac{AA_{1}}{A_{1}C} =2$. Let $AB = 6x \Rightarrow AP = 4x, PB = 2x$. Let $M$ be the point on $AB$ such that $CM \| B_1P$. Since $\frac{BP}{PM} = \frac{BB_{1}}{B_{1}C} = 2$ and $PB = 2x$ we got that $PM = x, BM = 3x = \frac{6x}{2} \Rightarrow M$ is the midpoint of $AB$ and then $CM \perp AB \Rightarrow B_{1}P \perp AB$ and since $CM$ is also a bisector of $\angle ACB$, $\angle PB_{1}B = \frac{\angle ACB}{2}$. But in cyclic quadrilateral $PQB_{1}B$ we got $\angle PQB = \angle PB_{1}B = \frac{\angle ACB}{2}$. Done!