Let $O', B', C'$ be the reflexion of $O,B,C$ w.r.t $EF$. Since $P$ and $A$ are symmetric to each other w.r.t $EF$, $O'$ is the circumcenter of $\triangle PB'C'$. By symmetry, $BC \cap B'C' = T$ and $BB'C'C$ is an isosceles trapezoid, which is concyclic. Finally by the radical axis theorem, the radical axis of $\odot O'$, $\odot D$ and $\odot BB'C'C$ are concurrent, in particular at $T$, which finishes the proof since $P \in \odot O' \cap \odot D$.

Nice problem! Here is an unexpected inversion solution. Let $O'$ be the reflection of $O$ across $EF$ and $M$ be the midpoint of $AP$. Notice that lines $EF, FD, DE$, $OD, OE, OF$ are perpendicular to lines $AP, BP, CP$, $BC, CA, AB$ respectively. Let $\triangle P_aP_bP_c$ be the pedal triangle of $P$ w.r.t. $\triangle ABC$. Perform inversion at $P$ with arbitrary radius. Denote the inverse image by asterisk, we get the following. $AP\perp P_b^*P_c^*$ thus $\triangle P_a^*P_b^*P_c^* \cup P$ and $\triangle DEF\cup O$ are homothetic. $M^*$ is reflection of $P$ across $P_b^*P_c^*$. But $O'$ is reflection of $O$ across $EF$. Thus $DO\parallel M^*P_a^*$. $T^*$ is foot from $P$ to $M^*P_a^*$. Hence $PT\perp M^*P_a^* \parallel DO'$ so we are done.

TelvCohl wrote: Given a $ \triangle ABC $ and a point $ P. $ Let $ O, D, E, F $ be the circumcenter of $ \triangle ABC, \triangle BPC, \triangle CPA, \triangle APB, $ respectively and let $ T $ be the intersection of $ BC $ with $ EF. $ Prove that the reflection of $ O $ in $ EF $ lies on the perpendicular from $ D $ to $ PT. $ Proposed by Telv Cohl Hopefully correct. Let $O'$ be the reflection of $O$ in $EF$. Notice that $P$ is the reflection of $A$ in $EF$. We have, $TO'^2 + PD^2 = TO^2+BD^2$ and $TD^2+O'P^2=TD^2+AO^2=TD^2+BO^2$. Further, $OD\perp TB$, so $TO'^2 + PD^2 = TD^2+O'P^2$ which gives us $PT\perp O'D$ and we are done.

Nice and easy. Here's my solution: Note that it suffices to show that $T$ lies on the radical axis of $\odot (PBC) $ and $\odot (O', O'P)$. Using the fact that $OO'AP$ is an isosceles trapezoid, this translates to proving that $$TB \cdot TC=|O'T^2-O'P^2| \Longleftrightarrow TB \cdot TC=|OT^2-OA^2|$$which is true since both sides are equal to power of $T$ with respect to $\odot (ABC) $. Hence, done. $\blacksquare$

Let $O'$ be the reflection of $O$ in $EF$. Note that $EF$ is just the perpendicular bisector of $AP$. Let $AT$ intersect $(ABC)$ again at $X$, and reflect $X$ across $EF$ to $X' \in PT$. Then $O'P = OA = OX = O'X'$, and $TX' \cdot TP = TX \cdot TA = TB \cdot TC$ implies $X'$ lies on $(PBC)$. So $O'D$ is the perpendicular bisector of $X'P$, hence proved.