We have $\angle KBC = \angle KAC = \frac{1}{2}{\angle BAC} = \angle KAB = \angle KLB$. Thus, $KB$ is tangent to $(MLB)$. Similarly, $KC$ is tangent to $(MLC)$. Now, $KB^2 = KC^2 = KM \cdot KL = KP \cdot KQ$. Thus, $KB$ is tangent to $(BPQ)$ and $KC$ is tangent to $(CPQ)$. We have, $\angle KBQ = \angle KPB$ and $\angle KCQ = \angle KPC$. Similarly, $LB^2 = LC^2 = LM \cdot LK = LQ \cdot LR$. Thus, $LB$ is tangent to $(BQR)$ and $LC$ is tangent to $(CQR)$. We have $\angle LBQ = \angle LRB$ and $\angle LCQ = \angle LRC$. So, $\angle BPC + \angle BRC = (\angle KPB + \angle KPC) + (\angle LRB + \angle LRC)$ $\angle BPC + \angle BRC = \angle KBQ + \angle KCQ + \angle LBQ + \angle LCQ$ $= (\angle LBQ + \angle KBQ) + (\angle KCQ + \angle LCQ)$ $= \angle LBK + \angle LCK = 180^\circ$ So, $B, P, C, R$ are concyclic.

By spiral similarity, $\triangle{MPL} \sim \triangle{MKR}$, so $MR \cdot MP = MK \cdot ML = MB \cdot MC$. So if $R'$ is the reflection of $R$ over $KL$, then $BPCR'$ is cyclic by Power of a Point and $R$ lies on its circumcircle. $\square$

v4913 wrote: By spiral similarity, $\triangle{MPL} \sim \triangle{MKR}$, so $MR \cdot MP = MK \cdot ML = MB \cdot MC$. So if $R'$ is the reflection of $R$ over $KL$, then $BPCR'$ is cyclic by Power of a Point and $R$ lies on its circumcircle. $\square$ theres a sol with inversion along these lines