You can prove $\triangle ACD = \triangle AEF \implies AB$ is bisector of $\angle CBF$. So $AB , CM , FN$ are concyclic. Let $O$ is common point. Power of point th $OC \cdot OM = OA \cdot OB = OF \cdot ON$ which is equivalent to result.

Let $\mathcal{S}$ denote the spiral similarity such that $\mathcal{S}(C)=E$ and $\mathcal{S}(D)=F$. Applying the well-known lemma concerning the unique center of spiral similarity, the center of $\mathcal{S}$ is $A$ and, since $CD=EF$, $\mathcal{S}$ is a rotation. This implies that $AC=AE$ and that $AD=AF$. Therefore $A$ is the midpoint of arcs $\widehat{CAE}$ and $\widehat{DAF}$ of circles $\Gamma_1$ and $\Gamma_2$, respectively. Therefore $AB$ is the bisector of angle $\angle{CBF}$ and, since $M$ and $N$ are the midpoints of arcs $\widehat{PB}$ and $\widehat{QB}$, $CM$ and $FN$ are the bisectors of angles $\angle{BCF}$ and $\angle{BFC}$, respectively. Therefore $CM$, $FN$ and $AB$ concur at the incenter of triangle $\triangle{BCF}$ which, since $AB$ is the radical axis of $\Gamma_1$ and $\Gamma_2$, implies that $C$, $M$, $F$ and $N$ are concyclic.

At first I thought of performing an inversion with center $A$ but after a while found that it was not the case. Note that $\angle{AEC}=\angle{ABD}=\angle{AFD}$ and $\angle{ACE}=\angle{ABE}=\angle{ADF}$,so $\triangle{AEC} \sim \triangle{AFD} \implies \frac{AE}{AC}=\frac{AF}{AD} \implies \frac{AE}{AF}=\frac{AC}{AD}$.Also note that $\angle{EAF}=\angle{CAD}$.Thus $\triangle{AEF} \sim \triangle{ACD}$.But it is given that $CD=EF$ so in fact $\triangle{AEF} \cong \triangle{ACD}$.Thus $AE=AC=AF=AD$ or $\angle{ACE}=\angle{AEC}=\angle{AFD}=\angle{ADF}$.This in turn yeilds that $\angle{CBA}=\angle{FBA}$ or $AB$ bisects $\angle{CBF}$.Also since $M$ and $N$ are the midpoints of arcs $PB$ and $BQ$ they are the bisectors of $\angle{BCF}$ and $\angle{BFC}$.Combining these details we see that $AB,CM,FN$ concur at the incenter $I$ of $\triangle{BCF}$.Thus $IM \cdot IC=IB \cdot IA=IN \cdot IF$ or $MNFC$ is a cyclic quadrilateral.

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How can it be solved by inversion? i tried, but not works well...(or lack of experience)

sayantanchakraborty wrote: At first I thought of performing an inversion with center $A$ but after a while found that it was not the case. Note that $\angle{AEC}=\angle{ABD}=\angle{AFD}$ and $\angle{ACE}=\angle{ABE}=\angle{ADF}$,so $\triangle{AEC} \sim \triangle{AFD} \implies \frac{AE}{AC}=\frac{AF}{AD} \implies \frac{AE}{AF}=\frac{AC}{AD}$.Also note that $\angle{EAF}=\angle{CAD}$.Thus $\triangle{AEF} \sim \triangle{ACD}$.But it is given that $CD=EF$ so in fact $\triangle{AEF} \cong \triangle{ACD}$.Thus $AE=AC=AF=AD$ or $\angle{ACE}=\angle{AEC}=\angle{AFD}=\angle{ADF}$.This in turn yeilds that $\angle{CBA}=\angle{FBA}$ or $AB$ bisects $\angle{CBF}$.Also since $M$ and $N$ are the midpoints of arcs $PB$ and $BQ$ they are the bisectors of $\angle{BCF}$ and $\angle{BFC}$.Combining these details we see that $AB,CM,FN$ concur at the incenter $I$ of $\triangle{BCF}$.Thus $IM \cdot IC=IB \cdot IA=IN \cdot IF$ or $MNFC$ is a cyclic quadrilateral.

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Can you explain why OC.OM=OA.OB=OF.ON

Let $CM$ and $FN$ meet at $I$. clearly $I$ is incenter of $CBF$. it's well known that $AEF$ and $ACD$ are similar and we have $EF = CD$ so they're congruent so $\angle FBA = \angle FDA = \angle 90 - \angle \frac {FAD}{2} = \angle 90 - \angle \frac {FBD}{2} = \angle 90 - \angle \frac {EBC}{2} = \angle 90 - \angle \frac {EAC}{2} = \angle AEC = \angle ABC$ so $AB$ is angle bisector of $CBF$ so $I$ lies on $AB$ which is radical axis of circles so $CI.IM = AI.IB = FI.IN$ so $CFMN$ is cyclic. we're Done.