Let $R_1$ be the radius of $\triangle MQC$. Let $R_2$ be the radius of $\triangle NPB$. Using the fact that $NB=NI=NA$, $MC=MI=MA$, and since $NM \parallel PQ$, we get: $\angle PIB= \angle MIQ = \angle IMN= \angle NCA= \angle NAB= \angle NBA$ $\angle QIC= \angle NIP= \angle INM= \angle MBA= \angle MAC= \angle MCA$ Therefore, by trig ceva in $\triangle NBI$ and $\triangle MIC$, we get: $\frac{ \sin \angle BNP}{ \sin \angle PNI} = \frac{ \sin \angle IMQ}{ \sin \angle QMC}$ By Two Equal Angles Lemma we get $\angle BNP= \angle IMQ$ We now easily see that $\angle ANP+ \angle AMQ= 180$. And $AP= AQ$. By law of sines in $\triangle ANP$ and $\triangle AMQ$, we get: $\frac{NP}{ \sin \angle NAP} = \frac{AP}{ \sin \angle ANP} = \frac{AQ}{ \sin \angle AMQ} = \frac{MQ}{ \sin \angle MAQ}$. By law of sines in $\triangle MQC$ and $\triangle NPB$ we get: $2R_1= \frac{MQ}{ \sin \angle MCQ} = \frac{NP}{ \sin \angle NBP} = 2R_2$ $\Longrightarrow R_1=R_2$ Done!

Let $T = AI\cap\odot(ABC)$. It's well known that $AT\perp MN$, so $AI\perp PQ$. Recall that by Fact 5 $BN = NI$ and $CM = MI$; combined with the fact that $\angle BNC = \angle BMC$ we get that $\triangle BNI\sim\triangle ICM$. Furthermore, remark that \[\angle API = 90^\circ-\tfrac12A = 180^\circ - \left(90^\circ+\dfrac12\angle A\right) = 180^\circ - \angle BIC = \angle NIB.\]Thus $\angle PBI = \angle NIP$, so $\angle NBP=\angle PIB$. Similar reasoning works for the other triangle $\triangle CMI$, so in fact $P$ and $Q$ are corresponding points in $\triangle BNI$ and $\triangle IMC$ respectively. As such, \[\dfrac{R_{\odot(MQC)}}{R_{\odot(BPN)}} = \left(\dfrac{R_{\odot(MQC)}}{R_{\odot(MQI)}}\right)\left(\dfrac{CM}{CN}\right) = \left(\dfrac{\sin\angle MCA}{\sin\angle ABN}\right)\left(\dfrac{MA}{AN}\right) = 1,\]where the last equality is from Extended Law of Sines. Done. $\blacksquare$

1) $\angle NBP=\angle NAP$ And since $\triangle BNP$ shares $NP$ with $\triangle NPA$, they have both the same circumradius. Analogously $\triangle CQM$ and $\triangle AQM$ have the same circumradius. So its enough to prove $\triangle AQM$ and $\triangle NPA$ have the same circumradius. 2) It´s known that $NI=NA$ and $MI=MA \Rightarrow MN\perp AI\Rightarrow PQ\perp AI$ Since $AI$ its the angle bisector of $\angle BAC\Rightarrow AP=AQ$ So its enough to prove that $\angle ANP=180-\angle AMQ \Leftrightarrow NP\cap MQ$ is on $\Gamma$ 3) Consider $NP\cap \Gamma=D$ and $MD\cap AC=Q'$ From Pascal we get $AB\cap ND=P$ , $BM\cap CN=I$ and $AC\cap MD=Q'$ are collinear but $P, I, Q$ are collinear $\Rightarrow Q'=Q$ $Q.E.D$

My solution: Let $NP$ cut $MQ$ at $R$. From Pascal's theorem, we have $R$ lies on $(O)$. We have $\angle RNB=\angle RMB$ and $\angle NBP=\angle NMI=\angle MIQ$$\Longrightarrow $$\triangle NPB\sim \triangle MQI$$\Longrightarrow $$\frac {NP}{MQ}=\frac {NB}{MI}=\frac {NI}{MI}=\frac {sin\frac {\angle C}{2}}{sin\frac {\angle B}{2}}$ $\Longrightarrow $the circumradius of $\odot (BNP)$ and $\odot (CMQ)$ are equal

@doxuanlong1505 apart from some typos that you can rectify, nice solution.

Let $\odot (CMQ)$ $\cap$ $NC$ = $D$. $\angle NBP$ = $\angle$ $ICA$ . It´s easy to see: $AP$ = $AQ$ $\longrightarrow$ $PI$ = $IQ$. If prove $NP$ = $QD$ the circumradius of $\odot (BNP)$ and $\odot (CMQ)$ are equal. Let $NP$ $\cap$ $(O)$ = $E$. It´s easy to see $IPBE$ is cyclic. By angle chasing $IQCE$ is cyclic $\longrightarrow$ $E$, $Q$ and $M$ are collinear. Then: $\angle$ $PNI$ = $\angle$ $IDQ$. $\triangle$ $NPI$ $\cong$ $\triangle$ $IDQ$ $\longrightarrow$ $NP$ = $QD$, done.

Math_CYCR wrote: Let $R_1$ be the radius of $\triangle MQC$. Let $R_2$ be the radius of $\triangle NPB$. Using the fact that $NB=NI=NA$, $MC=MI=MA$, and since $NM \parallel PQ$, we get: $\angle PIB= \angle MIQ = \angle IMN= \angle NCA= \angle NAB= \angle NBA$ $\angle QIC= \angle NIP= \angle INM= \angle MBA= \angle MAC= \angle MCA$ Therefore, by trig ceva in $\triangle NBI$ and $\triangle MIC$, we get: $\frac{ \sin \angle BNP}{ \sin \angle PNI} = \frac{ \sin \angle IMQ}{ \sin \angle QMC}$ By Two Equal Angles Lemma we get $\angle BNP= \angle IMQ$ We now easily see that $\angle ANP+ \angle AMQ= 180$. And $AP= AQ$. By law of sines in $\triangle ANP$ and $\triangle AMQ$, we get: $\frac{NP}{ \sin \angle NAP} = \frac{AP}{ \sin \angle ANP} = \frac{AQ}{ \sin \angle AMQ} = \frac{MQ}{ \sin \angle MAQ}$. By law of sines in $\triangle MQC$ and $\triangle NPB$ we get: $2R_1= \frac{MQ}{ \sin \angle MCQ} = \frac{NP}{ \sin \angle NBP} = 2R_2$ $\Longrightarrow R_1=R_2$ Done! OMG MY FRIEND, I THOUGHT SAME THING! My solution is identical, so it would be useless post my solution...

It's easy to infer that $AI$ bisects $\overline{PQ}$. Let $E$ be the second intersection point of $BI$ and $(CQM$) and $D$ the second intersection of $CI$ and $(CQM)$. Note that $$\angle IEQ=\angle MCA=\angle IBP\therefore PB\parallel EQ$$so $BI=IE$ and hence $BPEQ$ is a parallelogram, thus $EQ=PB$. Observe that $$\angle IED=\angle MCI=\angle NBI\therefore NB\parallel ED$$therefore $\frac{IN}{ID}=\frac{BI}{IE}=1$ which implies $NI=ID$, then $NB=ED$. Finally, since $DN$ and $PQ$ bisect each other at $I$, so $NQDP$ is a parallelogram with $NP=QD$, then $\bigtriangleup NPB\cong \bigtriangleup DQE$, which gives the required result.