Solution:

Extend line $PC$ through $C$ to point $D$ such that $CD = BP$. Note that $\angle ACD = \pi - \angle PCA = \angle ABP$ (since quadrilateral $ABPC$ is cyclic), and $AC = AB$ since $\triangle ABC$ is equilateral. Consequently, $\triangle ACD \cong \triangle ABP$. In particular, we have $\angle PDA = \angle CDA = \angle BPA = \angle BCA$ (again by cyclicity) $= \pi/3$. But also $\angle APD = \angle APC = \angle ABC$ (cyclicity) $= \pi/3$. We conclude that triangle $APD$ is equilateral. So, $PA = PD = PC + CD = PC + BP$.