, 2024

Let $ω$ be the circumcircle of triangle $AEF$ and let $I_A$ be the A-excenter of triangle $ABC$. First, we prove that $I_A$ is the midpoint of the arc $EF$ of $ω$ that does not contain point $A$ (see the left figure).
To that end, note that since $I_A$ lies on the external angle bisector of $∠B$ and $BE = BD$, triangles $BEI_A$ and $BDI_A$ are congruent (SAS). Similarly, triangles $CFI_A$ and $CDI_A$ are congruent, so in particular $I_A E = I_A F$. Moreover, $∠BEI_A + ∠CFI_A = ∠BDI_A + ∠CDI_A = 180^\circ$, hence the points $A, E, I_A, F$ lie on a single circle in this order.

Next, we prove that $P$ is the second intersection of $I_A D$ and $ω$ (see the middle figure). Let $I$ be the incenter of triangle $ABC$. Then $ED \parallel BI$ and $DF \parallel IC$. Setting $∠EAF = α$, we get $∠EDF = ∠BIC = 90^\circ + \frac{1}{2}α$, thus $∠EPF = α = ∠EAF$, so $P$ lies on $ω$. Since $I_A$ is the midpoint of arc, it lies on the angle bisector $PD$, so $P$ lies both on $I_A D$ and on $ω$ as claimed.
Finally, we show that $P$ lies on that arc of $ω$ which lies inside $Ω$ (see the right figure). Let $M \neq A$ be the second intersection of $ω$ and $Ω$ (if they are tangent, we set $M = A$). Then $M$ is the center of the spiral similarity that maps $BE$ to $CF$ (alternatively, we angle-chase that triangles $MBE$ and $MCF$ are similar by AA). Thus $MB/MC = BE/CF = BD/DC$, so $MD$ is the angle bisector of $BMC$, and so it passes through the midpoint $S$ of the arc $BC$ of $Ω$ that does not contain $A$.
Now forget about points $B, C, E, F$ and focus on circles $Ω, ω$ and on the points $A, M, I_A, S, D, P$. Circles $Ω$ and $ω$ share points $A$ and $M$. Being the A-excenter of $ABC$, point $I_A$ belongs to that arc $AM$ of $ω$ which lies outside of $Ω$ (e.g. since $AI_A > AS$). Point $S$ lies on the segment $AI_A$ and point $D$ lies on the segment $SM$, so point $D$ lies inside the angle $AI_A M$. Thus, point $P = I_A D \cap ω$ belongs to the other arc $AM$ of $ω$ than $I_A$, namely to the one which lies inside $Ω$.