Let $O$ be the center of $\omega$ and let $V$ be the intersection of $M P$ with $A K$. We first prove that $\angle P V L=\angle P O L$. If $V$ and $O$ coincide, there is nothing to prove. If $V$ and $O$ do not coincide, then $\angle O V P=90^{\circ}=\angle O L P$, so $O V P L$ or $V O P L$ is a cyclic quadrilateral. (In fact, $Q$ also lies on the corresponding circumscribed circle.) From this, it follows that $\angle P V L=\angle P O L$. We now have in all cases $\angle M V L=\angle P V L=\angle P O L$. Since $P L$ and $P Q$ are tangents to $\omega$, $\triangle O Q P \cong \triangle O L P$, so $\angle P O L=\frac{1}{2} \angle Q O L$. By the central angle theorem applied to $\omega$, this angle is also equal to $\angle Q A L$. Altogether, we find

$$\angle M V L=\angle P O L=\angle Q A L=\angle M A L$$

which implies that $M V A L$ is a cyclic quadrilateral. Therefore, $\angle A L M=180^{\circ}-\angle A V M=90^{\circ}$. Furthermore, by Thales' theorem, $\angle A L K=90^{\circ}$, so $\angle A L M=\angle A L K$, which means that $L$, $M$, and $K$ lie on a straight line.