On side $AB$ of triangle $ABC$, point $M$ is selected. A straight line passing through $M$ intersects the segment $AC$ at point $N$ and the ray $CB$ at point $K$. The circumscribed circle of the triangle $AMN$ intersects $\omega$, the circumscribed circle of the triangle $ABC$, at points $A$ and $S$. Straight lines $SM$ and $SK$ intersect with $\omega$ for the second time at points $P$ and $Q$, respectively. Prove that $AC = PQ$.