I feel nostalgic now, I was a Cuban contestant back in 2003 and I managed to solve this problem, here is my solution. Let $R$ be the second intersection point of $PB$ with $\omega$ and $M=ER\cap AP$. 1- $\angle ABE = 180-\angle ERB=\angle ERP$, on the other hand $\angle EPB=\angle EAB$, then $\triangle EBA\sim \triangle ERP$ and then $\angle REP=\angle BEA=\angle BPA$. 2- from the previous $MP$ is tangent to $(EPR)$ at $P$, then $MP^2=MR\cdot ME=MC^2$, so $M$ is the midpoint of $PC$. 3- Now, $\angle MQP = \angle MPQ=\angle MEP$ and then $EPMQ$ is cyclic and $\angle MEQ=\angle MPQ$ and the result follows.