Solution:

Let $M$ be the midpoint of $A B$ and let $N$ be the center of $\omega$. Then $M$ is the circumcenter of triangle $A B C$, so points $M, N$ and $R$ are collinear. From $Q N \parallel A M$ we get $\angle A M R=\angle Q N R$. Besides that, triangles $A M R$ and $Q N R$ are isosceles, therefore $\angle M R A=\angle N R Q$; thus points $A, Q, R$ are collinear.

Right angled triangles $A F Q$ and $A R B$ are similar, which implies $\frac{A Q}{A B}=\frac{A F}{A R}$, that is $A Q \cdot A R=A F \cdot A B$. The power of point $A$ with respect to $\omega$ gives $A Q \cdot A R=A P^{2}$. Also, from similar triangles $A B C$ and $A C F$ we get $A F \cdot A B= A C^{2}$. Now, the claim follows from $A C^{2}=A F \cdot A B=A Q \cdot A R=A P^{2}$.