BALKAN-SL 2019 G5

Let $A B C(B C>A C)$ be an acute triangle with circumcircle $k$ centered at $O$. The tangent to $k$ at $C$ intersects the line $A B$ at the point $D$. The circumcircles of triangles $B C D, O C D$ and $A O B$ intersect the ray $C A$ (beyond $A$ ) at the points $Q, P$ and $K$, respectively, such that $P \in(A K)$ and $K \in(P Q)$. The line $P D$ intersects the circumcircle of triangle $B K Q$ at the point $T$, so that $P$ and $T$ are in different halfplanes with respect to $B Q$. Prove that $T B=T Q$.