This is well known. Annyways As $\angle AOC+\angle BOD=\pi$. So there $\exists$ an Isogonal Conjuage of $O$ WRT $ABCD$. So, $M,N,P,Q$ are concyclic and trivially $M,N,P,Q,M'N',P',Q'$ lies on that circle.

1)$\overline{CO}^2=\overline{CN}*\overline{CB}=\overline{CP}*\overline{CD}$ gives that (1)$B, N, P, D$ concyclic. Similarly (2)$C, P, Q, A$ concyclic. Then we have $$\angle NMQ= \angle NMO+ \angle QMO=\angle OBN+\angle OAQ$$From (1) we get that $\angle NPO=90^{\circ}-\angle OBN$ and from (2) we get $\angle QPO=90^{\circ}-\angle QAO$ from which $$\angle NPO+\angle QPO=180^{\circ}-(\angle OBN+\angle QAO)=180^{\circ}-\angle NMQ$$which implies that $M,N,P,Q$ concyclic. 2)We will prove that $N, Q', P, Q$ concyclic. Similarly we get that all of them concyclic. $$\angle Q'QP=\angle OQP=\angle ODP=\angle PNC=\angle PNQ'$$We are done! (The last equality follows from (1))