Beginners' Competition

$BCDE$ 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 $\overline{PA}=\overline{PD}$.