Solution:

Note that $\angle BPA = 180^{\circ} - 76^{\circ} - 52^{\circ} = 52^{\circ}$. Since $\angle PBA = 52^{\circ}$, then $\triangle BPA$ is isosceles and $|AB| = |AP| = |PD|$. But $|AB| = |CD|$, so by transitivity $|PD| = |CD|$ and therefore $\triangle PCD$ is also isosceles. Since $\angle CDA = 180^{\circ} - 76^{\circ} = 104^{\circ}$, then $\angle PCD = \frac{180^{\circ} - 104^{\circ}}{2} = 38^{\circ}$.