Solution:

The interior angles of a regular $n$-gon measure $\frac{n-2}{n} 180^{\circ}$. Hence the interior angles of a regular pentagon measure $\frac{3}{5} 180^{\circ}=108^{\circ}$. The triangles $EAB$ and $CBA$ are isosceles, so $\widehat{AEB}=\widehat{EBA}=\widehat{BCA}=\widehat{CAB}=\frac{180^{\circ}-108^{\circ}}{2}=36^{\circ}$. Then $\widehat{APB}=180^{\circ}-\widehat{PAB}-\widehat{PBA}=108^{\circ}$, $\widehat{CPB}=180^{\circ}-\widehat{APB}=72^{\circ}$ and $\widehat{CBP}=\widehat{CBA}-\widehat{PBA}=72^{\circ}$. Hence triangle $CBP$ is isosceles and $CP=CB=1$.

Solution 2:

By symmetry, the diagonal $AC$ is parallel to the side $DE$ and the diagonal $BE$ is parallel to the side $CD$; it follows that the quadrilateral $EPCD$ is a parallelogram and hence that $PC=ED=1$.