Solution:

Consider the homothety centered at $N$ that maps the circle $k$ to the circle $BCN$; let it map the point $P$ to $P_{1}$. The tangent to the circle $BCN$ at $P_{1}$ is parallel to the tangent to $k$ at $P$, i.e. to the line $BC$, which means that $P_{1}$ is the midpoint of the arc $BC$ of the circle $BCN$. Hence, $NP$ is the bisector of the angle $CNB$, so $\frac{B N}{C N}=\frac{B P}{C P}$. Moreover, by Menelaus's theorem we have $\frac{B M}{M C}=\frac{B R}{R A} \cdot \frac{A Q}{Q C}=\frac{B P}{P C}=\frac{B N}{N C}$, so $NM$ is the external bisector of the angle $CNB$.

Therefore, $N$ lies on the circle with diameter $MP$, and $L$ is the foot of the perpendicular from $M$ to $AP$. It remains to prove that $MS \perp AP$.

Let $L^{\prime}$ be the foot of the perpendicular from $S$ to $AP$. The points $A, L^{\prime}, Q, R, S$ lie on a circle $\omega$ with diameter $AS$. Inversion with respect to $k$ maps the circles $\omega$ and $SPL^{\prime}$ to the lines $QR$ and $BC$, respectively, so it maps the point $L^{\prime}$ to $M$. Therefore, $M$ lies on the line $SL^{\prime}$, from which the claim follows (and $L^{\prime} \equiv L$).