My solution: Let $ T $ be the midpoint of $ MP $ . From $ (M,P;B,C)=-1 \Longrightarrow TP^2=TB \cdot TC $ , so $ T $ is the radical center of $ \{ \odot (S), \odot (BNC), \odot (ABC) \} \Longrightarrow NT $ is the tangent of $ \odot (S) $ , hence we get $ TN=TP=TM \Longrightarrow MP $ is the diameter of $ \odot (MNP) \Longrightarrow ML \perp AP $ . ... $ (\star) $ Since $ M $ lie on the polar $ QR $ of $ A $ WRT $ \odot (S) $ , so $ AP $ is the polar of $ M $ WRT $ \odot (S) \Longrightarrow MS \perp AP $ , hence combine with $ (\star) $ we get $ S, L, M $ are collinear . Q.E.D ____________________________________________________________ P.S. For more property about this configuration you can see : http://www.artofproblemsolving.com/community/c6h17323 http://www.artofproblemsolving.com/community/c6h332584

Hello!Tevl Cohl was way faster than me.My solution is almost the same.Let me note that another approach for the relation $ML\perp AP$ is to use the lemma that $NP$ bisects $\angle{BNC}$ which,since $M,B,P,Q$ are a harmonic series,gives that $MN\perp NP$ as internal and external bisectors of the angle $\angle{BNC}$.

Posted before at http://www.artofproblemsolving.com/community/c6h318871