Let ,
$\angle CPY = \angle CBY = \angle p$
$\angle PBC = \angle PYC = \angle q$
$CX = CY \Rightarrow \angle CYX = \angle CXY = \angle q$
$AB = CY , BY = BY , \angle BAY = \angle BCY$
So,$\triangle ABY \cong \triangle BCY$
Thus, $\angle CBY = \angle AYB = \angle ACB = \angle p$
From $\triangle PXC$, $\angle p + \angle PCX = 180 - \angle PXC$
But, $\angle YXC = 180 - \angle PXC = \angle q$.
$\angle p + \angle PCX = \angle q \Rightarrow \angle PCB = \angle PBC$
$\therefore PB = PC$