$BCDE$ is a square since $\overline{BC} = \overline{CD} = \overline{DE}$ and $BC \perp CD$, $CD \perp DE$. Hence also the length of the segment $BE$ coincides with the side length of the pentagon $ABCDE$ and we have $BC \perp BE$ and $BE \perp DE$. Furthermore $\overline{AB} = \overline{AE} = \overline{BE}$, hence $ABE$ is an equilateral triangle. Now we have
$$\angle CBA = \angle CBEB + \angle EBA = 90^\circ + 60^\circ = 150^\circ,$$
$$\angle AED = \angle AEB + \angle BED = 60^\circ + 90^\circ = 150^\circ.$$
Since $\overline{AB} = \overline{BC} = \overline{DE} = \overline{EA}$ the isosceles triangles $ABC$ and $AED$ are congruent. We get
$$\angle BAC = \angle ACB = \angle DAE = \angle EDA = \frac{180^\circ - 150^\circ}{2} = 15^\circ.$$
Since every diagonal in a square bisects the right angles in its endpoints, we have
$$\angle ADP = \angle EDB - \angle EDA = 45^\circ - 15^\circ = 30^\circ.$$
$$\angle PAD = \angle BAE - \angle BAC - \angle DAE = 60^\circ - 2 \cdot 15^\circ = 30^\circ.$$
Hence $ADP$ is a isosceles triangle with basis $AD$ and it follows that $\overline{PA} = \overline{PD}$.