We claim that $MPQ$ is congruent to the triangle $DEF$ of points where the incircle of $ABC$ touches the sides. Indeed, we claim that upon reflection over the midpoint of $AB$, one triangle is mapped to the other. It is clear that $M$ is mapped to $F$. By symmetry, it suffices to prove that $P$ is mapped to $D$. But this is equivalent to $APM$ being similar to $BDF$. Both are iscosceles and have \[\angle PAM=2\angle PNM=2\angle PNC-2\angle MNA=180^\circ-\gamma-2\left(90^\circ-\angle NAB\right)=2\angle NAB-\gamma=180^\circ-\alpha-\gamma=\beta\]so they are indeed similar. Done!

Let line $AP$ and $BQ$ intersects at $R$.And incircle of $\triangle ABC$ tangent to sides $AB,BC,CA$ at$C_1,A_1,B_1$ respectively. We have $AN=AP$ the $\angle ANP=\angle APN=\angle CNK$ and because $CN=CK$ then we get $\angle NAP=\angle NCK$.That means $AP \parallel CK => AR \parallel CB$ similarly $BR \parallel AC$.That means $ABCR$ is parallelogramm.And we have well known $AM=BC_1$.That implies $BA_1=BC_1=AM=AN=AP$ Then because $ABCR$ is parallelogramm and $\angle BAP=\angle CBC_1$ and $AM=AP=BC_1=BA_1$ means $A_1C_1=MP$ Similarly $MQ=B_1C_1$ and PQ=A_1B_1 That means triangle PRQ and A_1B_1C_1 are equal.then their circumradiuses are equal