Notice that \[\measuredangle FAE= \measuredangle BAE= \measuredangle BDE= \measuredangle PDE= \measuredangle PFE\]and similarly $\measuredangle FAE= \measuredangle FEP$ implying $PE,PF$ are tangents to $AEF$ so line $AP$ is a symmedian as desired

Let $N\in EF$ such that $\angle NAE=\angle BAP$. We will prove that $NE=NF$. Let $\angle BAP=\alpha=\angle NAE$ and $\angle AEN=\theta$. $NE=NA.\frac{sin \alpha}{sin \theta}$ and $NF=NA.\frac{sin A+\alpha}{sin A-\theta}$ We have $PE=PF$ since $\angle EFP=\angle EDC=\angle A=\angle PEF$. By angle chasing we get $\angle PEA=\angle PEF-\angle AEF=\angle A-\theta$ and $\angle EAP=\angle A+\alpha$. Thus, \[\frac{sin \alpha}{sin \theta}=\frac{PF}{PA}=\frac{PE}{PA}=\frac{sin A+\alpha}{sin A-\theta}\]Which gives that $NE=NF$ as desired.$\blacksquare$