A tricky geometry problem I already have my own solution for the following question. But I am still interested in other elegant solutions without trigonometry if possible. 





This is my own solution. I am lazy to upload the TeX code, I am sorry.
A hint:

Draw a regular $18$-gon $Q$. It has the property that the angle between neigbouring diagonals emanating from the same vertex is $10^\circ$.

The points $B$, $C$, $D$ of your figure can be realized as vertices of $Q$; furthermore the line $C\vee P$ is  a diagonal emanating from $C$, and $D\vee A$, $\ D\vee E$  are diagonals emanating from $D$.

I think that the solution of your problem is hidden in this figure. The nontrivial point is the fact that the line $B\vee P$ is also a diagonal, i.e, that three diagonals of $Q$ meet at  $P$. This in turn has to do with algebraic relations among the $18$th roots of unity.
Construct equilateral triangle PDF (where F is below BC)

Join FB and FC

Triangles FPC and DPC are congruent (SAS)

Therefore CF = CD

angle CFD = angle CDF = 20 deg.

Then C,F,B, D concylic.

angle FBC = angle FDC = 20 deg

Triangle BPD and BFD are congruent (SAS)

Therefore angle PBD = 40 deg.