Invert about $ F$. Let $ A,B,C,D,E,P$ be transformed into $ A^\prime,B^\prime,C^\prime,D^\prime,E^\prime,P^\prime$. $ AC$ is transformed into circle $ \Gamma_1=A^\prime C^\prime F$, and $ BC$ is transformed into circle $ \Gamma_2=B^\prime C^\prime F$. $ D^\prime$ and $ E^\prime$ are on $ \Gamma_2,\Gamma_1$ respectively so that $ D^\prime E^\prime$ is a common tangent. Since this tangent is the transformation of the incircle, it is parallel to $ A^\prime B^\prime$. Then $ P^\prime$ is the intersection of the radical axis $ FC^\prime$ and the common tangent $ D^\prime E^\prime$. Thus, $ P^\prime D^\prime=P^\prime E^\prime$. Since $ ABPE$ is cyclic, so is $ A^\prime B^\prime P^\prime E^\prime$. Since $ P^\prime E^\prime||A^\prime B^\prime$, $ A^\prime B^\prime P^\prime E^\prime$ is an iscoceles trapazoid. Since $ D^\prime E^\prime$, the common tangent is parallel to $ A^\prime B^\prime$, it follows that the projection of $ E^\prime$ onto $ A^\prime F$ bisects the segment $ A^\prime F$. Thus, $ D^\prime P^\prime=P^\prime E^\prime=A^\prime B^\prime-A^\prime F=B^\prime F$. Since the projection of $ D^\prime$ onto $ B^\prime F$ bisects the segment, it follows that the projection of $ F$ onto $ P^\prime D^\prime$ bisects the segment. Thus, the circle $ D^\prime P^\prime F$ is tangent to $ A^\prime B^\prime$ and thus $ DP$ is parallel to $ AB$, as desired.

$DE\bigcap AB={M}$ Then MP is the tangent of cyclic (PEFD). $PM\bigcap CA={K}, PB\bigcap DE={H}$ ABPE is cyclic then $\angle PBM=\angle KEP=\angle PDE=\angle KPE$ Hence PDBA is cyclic. Then $\angle BDE=\angle BPM=\angle DEA$ KPHE is cyclic. Hence $\angle PHD=\angle PKE \Rightarrow \angle HPD=\angle HDP$ Hence We have $ PD\left | \right |AB$

It suffices to prove that $\frac{CP}{CD} = \frac{PF}{BF}$ since $\measuredangle ABP = \measuredangle PEC = \measuredangle PFE$ and $\measuredangle BPF = 180 - \measuredangle BAC - \measuredangle FPE = 180 - \measuredangle BAC - \measuredangle AFE = \measuredangle AEF = \measuredangle FPE = \measuredangle BFX$ so $FX = FE$ and $\triangle PEF \sim \triangle FXB$ hence $$\frac{PF}{BF} = \frac{PE}{FX} = \frac{PE}{FE} = \frac{Sin \measuredangle EFP}{Sin \measuredangle EPF} = \frac{Sin \measuredangle PEC}{Sin \measuredangle EPC} = \frac{CP}{CE} = \frac{CP}{CD}$$therefore we are done. $\blacksquare$