Here is a sketch. Obviously $ANCP$ and $BMCQ$ are cyclic. Extend $MN$ to meet $(ACP)$ and $(BCQ)$ at $X, Y$. The angle conditions imply that $ACPX$ and $BCQY$ are isosceles trapezoids. If we prove that $XPQY$ is cyclic, we will be done, as $O$ would be its center ($O$ lies on the perpendicular bisectors of $XP$ and $YQ$ using the isosceles trapezoids). Notice that $P-C-Q$ by an easy angle chase. To finish, note that $\angle YXP= \angle NCQ= 180- \angle YQP$, so we are done.

Spent 1.5 hours on this, then realized its an easy complex bash (only 1 page, with big handwriting )

Note that $N$ and $M$ are the center of the spiral similarity that sends $MC$ to $AP$ and $NC$ to $BQ$, respectively, therefore we have $$\triangle NMC\sim \triangle NAP\hspace{1cm} \triangle NMA\sim \triangle NCP \hspace{1cm}P\in\left(ACN\right)$$$$\triangle MNC\sim \triangle MBQ\hspace{1cm} \triangle MNB\sim \triangle MCQ \hspace{1cm}Q\in\left(BCM\right)$$ Lemma 1 $P, C, Q$ are collinear Proof: By $APCN$ cyclic and $\triangle NMC\sim \triangle NAP$ $$ \angle PCA=\angle PNA=CNM$$By $BMCQ$ cyclic and $\triangle MNB\sim \triangle MCQ$ $$\angle BMQ=\angle MCQ=\angle MNB$$ Therefore $\angle MCQ+\angle PCA=\angle MNB+\angle CNM=180^\circ$ $\hspace{20cm} \blacksquare$ Lemma 2 If $R=PQ\cap \left(ABC\right)$, then $A$ is the center of the spiral similarity that sends $PR$ to $NB$ Proof: By $ARCB$ and $APCN$ cyclic, $$\angle PRA=\angle NBA\hspace{1cm}\angle APR=\angle ANB$$$ \hspace{20cm} \blacksquare$ Lemma 3 $CQ=PR$ Proof: By $\triangle MNB\sim \triangle MCQ$ $$\frac{CQ}{MC}=\frac{NB}{MN}$$ And by Lemma 2 we have $\triangle APR\sim \triangle ANB$ and $\triangle ARB\sim \triangle APN\sim MCN$ $$\frac{PR}{BN}=\frac{AR}{AB}=\frac{AP}{AN}$$Therefore, $$\frac{PR}{CQ}=\frac{MN}{MC}\cdot \frac{AP}{AN}=1$$$\hspace{20cm} \blacksquare$ Finally by PoP from $P$ and $Q$ to $(ABC)$ $$OP^2=PR\cdot PC=PR\cdot (PR+RC)=QC\cdot QR=OQ^2$$$\hspace{20cm}\blacksquare$

Attachments:

Very nice problem Let, $\angle MCN= \alpha, \angle MNC=\phi, \angle NMC= \beta$ So $\alpha + \beta + \phi = 180^{\circ}$ Claim 1: A,N,C,P are concyclic and B,M,C,Q are concyclic. Proof: Since $\triangle MNC\sim \triangle ANP$, so $\angle MCN=\angle APN = \alpha$ $\rightarrow$ A,N,C,P are concyclic. $\square$ Since $\triangle MNC\sim \triangle MBQ$, so $\angle MCN=\angle MQB = \alpha$ $\rightarrow$ B,M,C,Q are concyclic. $\square$ Claim 2: Q,C,P are collinear Proof: Let's see what By Claim 1 $\angle ANP = \angle ACP = \phi$ By Claim 1 $\angle BCQ = \angle BMQ = \beta$ In $C$, $\angle ACP + \angle MCN + \angle BCQ = \alpha + \beta + \phi = 180^{\circ}$, so Q,C,P are collinear $\square$. Let $R: PQ \cap \odot ABC$ So $\angle BAR = \angle BCR = \beta, \angle ACB = \angle MCN = \angle ARB = \alpha$ So $\triangle ABR \sim \triangle MNC$ Let's see what $\triangle BMN \sim \triangle QMC$, and $\triangle ABN \sim \triangle ARP$ So $\triangle MCN \sim \triangle ARB$ Claim 3: $CQ = PR$ Proof: By $\triangle BMN \sim \triangle QMC$ \[\frac{CQ}{MC}= \frac{BN}{MN}\]By $\triangle ABN \sim \triangle ARP$ \[\frac{RP}{AR}=\frac{BN}{AB}\]Now \[\frac{CQ}{RP}=\frac{BN \cdot MC \cdot AB}{MN \cdot BN \cdot AR}=\frac{MC \cdot AB}{MN\cdot AR}=1\] So $CQ=RP$ $\square$ Finishing: Since $O$ belongs to the bisector of $RC$, but since $CQ= RP$ then $CR$ and $PQ$ have the same midpoint, then since $O$ is equidistant from $R$ and $C$ then $ O$ also equidistant from $P$ and $Q$, finally $OP=OQ$, as desired.

Let's present a different approach. As proven above we have that ANCP and BMCQ are cyclic. And P, C, and Q are collinear. Also by similarities given we have that $\bigtriangleup MCQ\sim\bigtriangleup MNB$ and $\bigtriangleup NCP\sim\bigtriangleup NMA$. By the Law of Cosines in $\bigtriangleup OCQ$ and $\bigtriangleup OCP$ it suffices that $OP^2=OQ^2 \iff CP^2-CQ^2=-2R\cdot CP\cos\angle OCP +2R\cdot CQ\cos\angle OCQ\iff |CP^2-CQ^2|=|(CP+CQ)(CP-CQ)|=2R|\cos\angle OCQ|(CP+CQ) \iff |CP-CQ|=2R|\cos\angle OCQ|$ where $R$ denotes the circumradius of $(ABC)$. Note that $\angle (MN,AB)=|90^{\circ}-\angle OCQ|$, hence $|\cos \angle OCQ|=\sin\angle (MN,AB)$. So it suffices to show that $|CP-CQ|=2R\sin\angle (MN,AB)$. From the similarities of triangles above, we get that $CP=\frac{NC}{NM}AM=\frac{\sin\angle CMN}{\sin\angle C}AM$ and similarly $CQ=\frac{\sin\angle CNM}{\sin C}BN$. So now it suffices to show that $|\frac{\sin\angle CMN}{\sin\angle (MN,AB)}AM - \frac{\sin\angle CNM}{\sin\angle (MN,AB)}BN|=2R\sin\angle C$. But note that $2R\sin\angle C=AB$. Let $J$ be the intersection of $MN$ and $AB$, then $|\frac{\sin\angle CMN}{\sin\angle (MN,AB)}AM - \frac{\sin\angle CNM}{\sin\angle (MN,AB)}BN| =| \frac{JA}{AM}AM-\frac{JB}{BN}BN|=|JA-JB|=|AB|=AB$, and we are done. I did not consider the $MN\parallel BC$, but it's not hard.