Let the tangents at $B, C$ meet at $Y$ and $AE \cap CQ=Z$. It's easy to see that $AC$ bisects $\angle BAE$, so $\angle BYC=180^{\circ}-\angle BAE$ and $ABYZ$ is cyclic. If $(PAB) \cap PQ=R$ is its Miquel point, it suffices to show that $D$ is its circumcenter. Since $AX=BC$ and $D$ is midpoint of the arc $ACB$, by spiral similarity we know that $DX=DC$. Moreover, $\triangle BAX$ and $\triangle YBC$ are congruent, we have that $BX=CY$. Since $\angle DXB=\alpha+\angle AXD=\angle BCY+\angle BCD=\angle DCY$, we obtain that $\triangle DXB$ and $\triangle DCY$ are congruent. Thus, $DA=DB=DY$ and indeed $D$ is the circumcenter of $ABYX$, as desired.