Let PI meet the circle of ABC at T. Observe that Angle AIC=120 degrees, so, BC1IA1 is cyclic. Now,by Reims on PBA1I and PBCT, line AI=line A1I is parallel to TC. Similarly, CI is parallel to AT. So, AICT is parallelogram. So, IT bisects AC.done.

Another approach is to write the condition "line $PI$ bisects the side $AC$" in sine form and then write some sine theorems.

Neat! starchan wrote: The bisectors $AA_1, CC_1$ of a triangle $ABC$ with $\angle B = 60^{\circ}$ meet at point $I$. The circumcircles of triangles $ABC, A_1IC_1$ meet at point $P$. Prove that the line $PI$ bisects the side $AC$. Note that $B, A_1, I, C_1$ are concyclic. Let $PI$ meet $(ABC)$ again at $Q$. Now $\angle BAQ=\angle BXI=\angle BA_1I=\angle A+\tfrac{1}{2}\angle C$ hence $\angle QAC = \tfrac{1}{2}\angle C = \angle ICA$. Likewise, $\angle QCA=\angle IAC$ hence $IAQC$ is a parallelogram or $PI$ bisects $AC$.