What is F ?

It seems like I did not include $F$ in my original post! I have posted a correction. Sorry for your inconvenience.

nice problem! It is well known that P is the Miquel point and thus the inverse of point F in circle O; thus inverting line BFD in circle O, we find that BPDO is cyclic, so if I is the intersection of line OPF and circle O, PBD = POD = IOD = 2IBD, which suffices to show that BI bisects PBD; everything else follows analagously and we are done.

let me expand on that well-known bit: basically, what we want to show is that P is the inverse of F in (O). Well, define Q as the inverse of F in (O) and we shall show that Q = P. Define G = BC ^ DE. Consider a tangent GT from G to (O), and consider the circle with center G passing through T. Suppose the inverse of O in (G) is X and let the inverse of G in (O) be G'. Then GX*GO = GT^2 = GO^2 - OT^2 = GO^2 - GO*OG' = GO*GG', so GX = GG' and thus X = G'. Thus, the pole of O in (G) is the line through X perpendicular to GO, which we know is the line through G' perpendicular to GO, which we know is the pole of G in (O), which we know is line AF. Thus F lies on the pole of O in (G), so O lies on the pole of F in (G), i.e. the inverse F' of F in (G) must be the foot of the perpendicular from O to GF. Thus GF' * GF = GT^2 = GB * GC, so F'BCF is cyclic. Now, invert this circle in (O) and see what circle it goes to. B and C stay fixed, and by definition, F goes to Q. As F' is the projection of O onto GF and GF is the pole of A in (O), F' must be the inverse of A, so F'DEF goes to AQBC, i.e. Q lies on the circumcircle of BC. But Q, being the inverse of F in (O), must also lie on ray OF. These two conditions clearly determine a unique point and as P also satisfies these conditions, Q = P, i.e. P is the inverse of F in (O).

I believe you are right! However, there are a lot of those who do not know inversion. Can anyone find an alternative solution?

Interesting...Let $BC \cap DE=G$ and redefine $P$ as the miquel point of $BEDC$, now from brokard notice that $O$ is orthocenter of $\triangle AFG$ and in addition $\angle APO=90=\angle GPO$ because by gliding principle there is a circle through $A$ and midpooints of $BE, CD$ that goes through $P$ and similarily for $G$, therefore $P$ lies in $AG$ and also by considering polars and orthocenter PoP (or invert) at we quickly check that $F,P$ are inverses in $(BEDC)$, so let $OF$ intersect $(BEDC)$ at $I,I'$ where $I$ is between $F,P$, now from invert we have that $-1=(P, F; I, I')$ and thus combined with $II'$ being diameter this gives that (from apollonious circle) $IE$ bisects $\angle PEC$ and $DI$ bisects $\angle PDB$ but also from invert we get $PEOC, PDOB$ cyclic which by mid-point of arc mean that $PI$ bisects $\angle EPC, \angle BPD$ therefore $I$ is the incenter of both $\triangle PBD$ and $\triangle PCE$, thus we are done .