Let $M \equiv AC \cap BD$. Note that $CP$ is the $C$- symmedian of triangle $DCB$. It is enough to show that $P$ is the foot of the $E$-symmedian of triangle $DEB$. This is equivalent to proving that $\frac{DC}{CB} = \frac{DE}{BE}$. It is almost trivial that cuadrilateral $ABED$ is harmonic since $AE$ passes through the midpoint of $BD$. This can be easily seen by the fact that $\frac{DE}{BE} = \frac{DM\cdot AB}{MB \cdot AD} = \frac{AB}{AD}$. Since $AB = DC$ and $AD = CB$ the conclusion follows.

Note: The quadrilateral isn't harmonic, since $AB*DE \not = AD*EB$ .

You are right. Anyways, my solution still holds.

can anyone tell that what is harmonic quadrilateral?

Dear Mathlinkers, John Casey has consacred a chapter to this question in Casey John, A Sequel to the first six books of the Elements of Euclid, Dublin, 1888. See on Google books or http://quod.lib.umich.edu/u/umhistmath?cginame=text-idx;id=navbarbrowselink;key=author;page=browse;value=c Sincerely Jean-Louis

Since quadrliateral $ABED$ is cyclic we have that $\angle DEA= \angle DBA$ and $\angle BAC=\angle BDE$. Denote the intersection of $BE$ and $DC$ as $K$. It is enough to prove that $\triangle BPE \sim \triangle BSD$, because that implies $\angle BEP=\angle BDS=\angle ABD=\angle AED$. It is easy to prove with some calculations. First denote angles $\angle CAD=\angle ACB=x, \angle BAC=\angle ACD=y, \angle DAC=\angle ACB=z, \angle DBA=\angle BDC= 180-(x+y+z).$ Expressing the sides of triangles $BPE$ and $BSD$ via the side $BC$ we get that $\frac{BE}{BP}= \frac{\sin{x}\sin{(y+z)}}{\sin{z}\sin{y}}(1)$ and $\frac{BD}{BS}=\frac{\sin{(x+y)}\sin{(y+z)}}{\sin{(x+y+z)}\sin{(x+y)}}(2).$ Also because diagonals of a parallelogram cut in half, we have that $\frac{\sin z}{\sin x}=\frac{\sin{(x+y+z)}}{\sin y}$ which along with $(1)$ and $(2)$ gives that $\frac{BE}{BP}=\frac{BD}{BS}\implies \triangle BPE \sim \triangle BSD$.$\blacksquare$

$\frac{DP}{PB}=\frac{CD^2}{CB^2}=\frac{AB^2}{AD^2}=\frac{\sin^2 \angle DAE}{\sin^2 \angle EAB}=\frac{BE^2}{DE^2}$ hence $PE$ is symmedian in $\Delta DEB$

From angles clearly $\triangle PBC \sim\triangle BEA\Rightarrow \frac{EB}{PB}=\frac{AB}{PC}=\frac{DC}{PC}\Rightarrow\triangle PBE \sim\triangle PCD$ and we're done.

Let's extend line $DE$ to intersect line $BC$ in point $F$. We have that $\angle PCF=\angle PDF \implies CDPF$ is cyclic quadrilateral. Quadrilateral $ABED$ is also cyclic. We have equal angles on the same arc, for example $\angle PFD=\angle PCD=\angle BCA=\angle DAE=\angle DBE \implies BPEF$ is also cyclic quadrilateral. We can use again same arc angles, so $\angle PEB=\angle PFB=\angle PDC=\angle DBA=\angle AED$. And that is all.

Since $\measuredangle PBE = \measuredangle DAE$, it's sufficient to show that $\Delta PEB \sim \Delta DEA$. Let $\measuredangle ACD = \measuredangle PCB = \measuredangle BAE = \alpha$ and $\measuredangle PCA = \beta$ Note that,$$\frac{PB}{AD} = \frac{PB}{BC} = \frac{sin \alpha }{sin \hat{BPC}}$$and $$\frac{BE}{AE} = \frac{sin \alpha}{sin \hat{EBA}}$$We can also see that $\measuredangle BPC =180- 2 \alpha - \beta - \measuredangle EBC$ and $\measuredangle EBA = 180- 2 \alpha - \beta - \measuredangle EBC$ So by the equalities above, we see that $$\frac{PB}{AD} = \frac{BE}{AE} \Rightarrow \frac{PB}{BE} = \frac{AD}{AE}$$and because $\measuredangle DAE = \measuredangle PBE = \alpha + \beta$, then $$\Delta PEB \sim \Delta DEA$$