We need prove that $ \angle BDF + \angle FBP=\angle BEP$. By radical center, we have $PF, BA, CE$ are concurrent at point $X$. $A$ is Miquel point of the triangle $CFX$. So, $PAEX$ is cyclic. We have $AB \parallel CD$. $\angle APE =\angle AXE= \angle ECD=\angle EAD$. So, $FD$ line tangent to $(PAE)$ circle. $$\angle BDF+\angle FBP=\angle BEA +\angle FAP=\angle BEA +\angle AEP=\angle BEP $$Done!

Splendid problem. I intended to give this at JBMO TST this year, but then Balkan MO 2024 came out and P1 had extremely similar solution ideas to the following approach. Without loss of generality let $P$ and $E$ lie on the same halfplane with respect to $AD$. Denote by $\ell$ the tangent to $B$ to the circumcircle of $BFD$ and let $K$ be a point on it which lies on different sides with $P$ with respect to $BD$. We have $\angle KBD = \angle BFD$, while $\ell$ is tangent to the circumcircle of $BEP$ if and only if $\angle KBE = \angle BPE$. Hence the desired tangency is equivalent to proving $\angle BPE = \angle BFD + \angle EBD$. (An alternative way to reduce to this equality is by introducing the centers of the two circles and seeing when are they collinear with $B$.) We compute $\angle BPE = \angle FPE - \angle FPB = (180^{\circ} - \angle FCE) - \angle FAB = 180^{\circ} - \angle BCE - \angle ADC$ (the second is from the cyclic $FPEC$ and $FPAB$, the third is from $AB \parallel CD$ due to $AD = BC$ in the circumcircle of $ABCDE$) and $\angle BFD = 180^{\circ} - 2\angle ADC$ (due to $AD = BC$ in the circumcircle of $ABCDE$). Therefore \[ \angle BPE - \angle BFD = \angle ADC - \angle BCE = \angle BCD - \angle BCE = \angle ECD = \angle EBD \]which completes the proof.

As in above solutions we prove $ \angle BDF + \angle FBP=\angle BEP.$ First note that $FP, AB$ and $EC$ concour at a point say $T.$ By Miquel points, $TEAP$ is cylic. Furthermore as, \[\angle CBT = \angle AET = \angle EPA = \angle BPA + \angle FPB = \angle BPA + \angle FAB = \angle BPA + \angle FBA = \angle BPA + \angle APT = \angle BPT\] we get $BC$ is tangent to $(BPT).$ Now to finish, \[\angle BEP = \angle BEA + \angle AEP = \angle BDA + \angle BTP = \angle BDF + \angle FBP.\]

Assassino9931 wrote: Splendid problem. I intended to give this at JBMO TST this year, but then Balkan MO 2024 came out and P1 had extremely similar solution ideas to the following approach. Without loss of generality let $P$ and $E$ lie on the same halfplane with respect to $AD$. Denote by $\ell$ the tangent to $B$ to the circumcircle of $BFD$ and let $K$ be a point on it which lies on different sides with $P$ with respect to $BD$. We have $\angle KBD = \angle BFD$, while $\ell$ is tangent to the circumcircle of $BEP$ if and only if $\angle KBE = \angle BPE$. Hence the desired tangency is equivalent to proving $\angle BPE = \angle BFD + \angle EBD$. (An alternative way to reduce to this equality is by introducing the centers of the two circles and seeing when are they collinear with $B$.) We compute $\angle BPE = \angle FPE - \angle FPB = (180^{\circ} - \angle FCE) - \angle FAB = 180^{\circ} - \angle BCE - \angle ADC$ (the second is from the cyclic $FPEC$ and $FPAB$, the third is from $AB \parallel CD$ due to $AD = BC$ in the circumcircle of $ABCDE$) and $\angle BFD = 180^{\circ} - 2\angle ADC$ (due to $AD = BC$ in the circumcircle of $ABCDE$). Therefore \[ \angle BPE - \angle BFD = \angle ADC - \angle BCE = \angle BCD - \angle BCE = \angle ECD = \angle EBD \]which completes the proof. i also solved with this method