3. A circle $\omega$ is circumscribed around triangle $ABC$. A line tangent to $\omega$ at point $C$ intersects the ray $BA$ at point $P$. On the ray $PC$ beyond point $C$, a point $Q$ is marked such that $PC = QC$. The segment $BQ$ intersects the circle $\omega$ again at point $K$. On the smaller arc $BK$ of the circle $\omega$, a point $L$ is marked such that $\angle LAK = \angle CQB$. Find the angle $\angle PCA$, given that $\angle ALQ = 60^{\circ}$.