Bump?????

Some progress: Let $E$ be the second intersection of $(AMN)$ and $(APQ)$. Then $KL$ is the perpendicular bisector of $AE$, so $RA = RE$, so it suffices to show that $RJ = RE$ or that $R$ is the circumcentre of $\triangle AJE$. Claim. The line $AC$ is the radical axis of circles $(AMN)$ and $APQ$. Proof. Let $CB$ intersect $(AMN)$ again at $X$ and $(APQ)$ again at $Y$. Since $CP = CN$, it suffices to show that $CX = CY$. Taking power of a point at $B$ with respect to $(AMN)$ we obtain that $BX = BA$. Likewise $DY = DA$. Thus \[ CX = BC - BX = BC - AB = CD - AD = CD - CY = CY \]as desired. Hence $A,C,E$ are collinear.