Let $ D$ be the reflection of $ B$ in $ O_1O_2$. Clearly, $ CD$ is a common tangent to $ C_1$ and $ C_2$. In particular, this means that $ \angle{O_2DC} = 90^{\circ}$ and hence $ PO_2DC$ is cyclic. Let $ \angle{ABP} = \alpha$. By symmetry we have that $ \angle{ABP} = \angle{CDP} = \alpha$. Since $ PO_2DC$ is cyclic it follows that $ PO_2C = \alpha$ and hence $ \angle{PCO_2} = 90^{\circ} - \alpha$. Finally, we see that $ \angle{O_2LB} = \angle{LBO_2} = 90^{\circ} - \alpha$. In conclusion $ \angle{PCO_2} = \angle{O_2LB}$, which implies that $ CL$ is tangent to $ C_2$.