Estonian Math Competitions

Note that the circles of the problem can be obtained from each other by homothetic transformation with center $B$ since both are tangent to sides $BA$ and $BC$. Let $X$ be the other intersection point of the line $BL$ with the circumcircle of the triangle $MPQ$.



The aforementioned homothety takes point $P$ to point $K$, point $M$ to point $L$, and point $X$ to point $M$. By the homothety, $\angle PXM=\angle KML$. Hence $\angle KML=\angle KLA=\angle LKM$. Finally, $$\begin{gathered} \angle PNM+\angle LNM=(180^{\circ }-\angle PXM)+\angle LKM \\ =(180^{\circ }-\angle KML)+\angle KML=180^{\circ }. \end{gathered}$$ This proves the desired claim that points $P$, $N$ and $L$ are collinear.