Školjka

Let $ABC$ be an acute-angled triangle with $|\angle{BAC}| > 45^{\circ}$ and with circumcentre $O$. The point $P$ lies in its interior such that the points $A$, $P$, $O$, $B$ lie on a circle and $BP$ is perpendicular to $CP$. The point $Q$ lies on the segment $BP$ such that $AQ$ is parallel to $PO$.\\\\ Prove that $|\angle{QCB}| = |\angle{PCO}|$.