Consider the inversion at $M$ with radius $MA$, and reflect wrt $AB$, call this transformation $\phi$. It suffices to prove that $\phi(P)=R$. Let $\phi(Q)=S$, and we have $\phi(K)=L$. Therefore $\measuredangle MLS=\measuredangle M\phi(S)\phi(L)=\measuredangle MQK=\measuredangle MQR=\measuredangle MLR$, so $LRS$ are collinear. Similarly, $\measuredangle MKS=\measuredangle MQL=\measuredangle MKP$, so $KPS$ are collinear, i.e. $M\phi(K)\phi(P)\phi(S)$ are concyclic, $\phi(P)$ lies on $(MLQ)$. Therefore $M$ is the Miquel point of quadrilateral $KPLR$, so $PLMS$ are concyclic, i.e. $\phi(P)\phi(L)\phi(S)$ are collinear, $\phi(P)$ is on $KQ$. We have $\phi(P)$ is the intersection of $KQ$ and $(MLQ)$, which is $R$, as desired.