Hints: Let $ \angle PAB=\alpha,\angle PAC=\beta$. Do some angle chasing. $ \angle CPK=\angle APK$ is straighforward. We need to prove that $ \angle MPA=\alpha+\beta$. Extend $ AP$ until it cuts $ BC$ at $ X$. $ BPX$ is similar to $ APB$, and $ \angle BPX=\alpha+\beta$. So we want to prove that $ PX$ is a median. Compare the area of CPX and PXB, and you'll get the desired result.

Raja Oktovin wrote: Let $ ABC$ be an isoceles triangle with $ AC = BC$. A point $ P$ lies inside $ ABC$ such that \[ \angle PAB = \angle PBC, \angle PAC = \angle PCB. \] Let $ M$ be the midpoint of $ AB$ and $ K$ be the intersection of $ BP$ and $ AC$. Prove that $ AP$ and $ PK$ trisect $ \angle MPC$. Let $ (O)$ is the circumcircle of $ \Delta APB$ then $ CA,CB$ are two tangents of $ (O)$. $ CP\cap (O)=\{J\}$ It's easy to see $ \angle CPK=\angle APK$ We need to prove that $ \angle CPK=\angle JPB=\angle APM$ or $ \angle APJ=\angle BPM$ Applying Ptoleme's theorem:$ AP.JB+AJ.PB=PJ.AB$ and $ AP.JB=PB.AJ$ we get $ AJ.PB=PJ.MB$ then $ \frac{AJ}{PJ}=\frac{MB}{PB}$ So $ \Delta AJP$~$ \Delta MBP$ Therefore $ \angle APJ=\angle MPB$

Let $ \angle PCB=\alpha ;\angle PBC=\beta $ $PA$ meets $BC$ at $N$ We have: $ \angle NCP =\angle NAC \rightarrow NC^2=NP.NA $ $ \angle NBP =\angle NBC \rightarrow NB^2=NP.NA $ Therefore ,$NC=NB$ otherwise , $N$ is the midpoint of $BC$ (*) It's easy to show that $ \triangle PAB \sim \triangle PBC$ and in addition , we have :$N$ is midpoint of AC So :$\angle PNB=angle MPA \rightarrow \angle PAB=\angle NPM=180-\alpha-\beta$ $= \angle CPK=\angle KPA$ Q.E.D

1) Easily, $\angle APK=\angle CPK \ (\ 1 \ )$ 2) $CP$ is symmedian of $\triangle ABP$, done! I thing it has been posted before, or, at least, similar! Best regards, sunken rock