Denote $M$= the Miquel point of $\triangle{ABC}$ and points $E,D,P$. $DP\parallel BB$,$EP\parallel AA$ therefore: $\angle{EPA}=\angle{BPD}=\angle{ACB}$ thus $\angle{EMA}=\angle{BMD}=\angle{ACB}$,but $\angle{DME}=180-\angle{ACB}$ so $M=AD\cap BE$ Let $N=$ midpoint of $BA$.It is well-known that $(CDE)$ and the circle with diameter $AB$ are orthogonal thus $NO'^2-CO'^2=NA^2$. $Lemma:$The locus of points $X$ for which:$UX^2-VX^2=constant$ is a line perpendicular on $UV$.($U,V$ are fixed points) So we only need to prove that $B',A'$ are on that line which is trivial: $NB'^2-B'C^2=B'A^2+AN^2-B'C^2=AN^2$ $NA'^2-A'C^2=BA'^2+BN^2-A'C^2=AN^2$

FWIW, an almost identical result was discussed in the thread of EGMO 2012 P1; see in particular sunken rock's conjecture and yetti's proof.

Let $A_1, B_1$ be points on $CB, CA$ respectively so that $\angle BAA_1=\angle CBB_1=\angle C$. Let $T$ be the center of a spiral similarity $A_1B \rightarrow AB_1$. Observe that $PD \perp BO \Longrightarrow PD \parallel AA_1,$ and similarly $PE \parallel BB_1$. It follows that the spiral similarity about $T$ sends $D$ to $E$. Since $\triangle DO'E$ has a fixed shape, $O'$ moves on a line. Looking at the cases $P=A$ and $P=B$ we conclude that $O'$ lies on the line $A'B'$. $\, \square$

Nice problem