First check, with directed angles, that \[\measuredangle PQO_2 = \measuredangle O_2AQ = 90^{\circ} + \measuredangle O_2AO_1 = 90^\circ+ \measuredangle O_1BO_2 = \measuredangle O_1BP.\]Now extend $\ell_2$ to hit $C_1$ again at $R$. The tangencies imply $\{BQ, BA\}$ antiparallel and $\{AB, AR\}$ antiparallel wrt $\angle BPA$. This means a reflection over the internal angle bisector of $\angle BPA$ followed by a homothety at $P$ will send $\triangle BAQ \mapsto \triangle ARB$, meaning the circumcenters of these triangles (i.e. $O_1$ and $O_2$) are isogonal wrt $\angle BPA$. Combining this with the angle equality we derived above yields the similar triangles we need.