We use barycentric coordinates on $\triangle ABC$. Let $BD=BU=u$, $CD=CE=a-u$, $FA=c-u$ and $AE=b+u-a$. Since $P\equiv BE\cap CF$ we get $$P=(u(a-u): (a-u)(c-u): u(b+u-a))$$ Then, the line $PD$ has equation $$ \begin{vmatrix} u(a-u) & (a-u)(c-u) & u(b+u-a) \\ 0 & a-u & u \\ x & y & z \end{vmatrix}= x(c+a-b-2u)-yu+z(a-u)=0$$For the concurrence of three lines $PD$, we taking three values variables of $u$: $(u_1,u_2,u_3)$. So we just need to show that $$ \begin{vmatrix} c+a-b-2u_1 & u_1 & a-u_1 \\ c+a-b-2u_2 & u_2 & a-u_2\\ c+a-b-2u_3 & u_3 & a-u_3 \end{vmatrix}=0$$which is true.

[asy][asy] size(10cm); defaultpen(fontsize(10pt)); pair A = dir(120), B = dir(210), C = dir(330), I = incenter(A,B,C), X = foot(I,B,C), D = (4B+C)/5, D1 = 2X-D, E = intersectionpoint(A--C, circle(C,abs(C-D))), F = intersectionpoint(A--B, circle(B,abs(B-D))), P = extension(B,E,C,F), A1 = 2X-A, Y = foot(I,C,A), Z = foot(I,A,B), E1 = 2Y-E, F1 = 2Z-F; draw(A--B--C--A, heavycyan); draw(circumcircle(D,E,F), purple); draw(P--D^^A--D1, heavygreen); draw(D--A1, heavygreen+dashed); draw(B--E^^C--F, heavycyan); draw(E--D--F^^E1--D1--F1, orange); draw(A--A1^^I--X, heavycyan+dotted); dot("$A$", A, dir(120)); dot("$B$", B, dir(210)); dot("$C$", C, dir(330)); dot("$D$", D, dir(230)); dot("$D'$", D1, dir(285)); dot("$E$", E, dir(60)); dot("$E'$", E1, dir(45)); dot("$F$", F, dir(180)); dot("$F'$", F1, dir(140)); dot("$X$", X, dir(270)); dot("$A'$", A1, dir(270)); dot("$P$", P, dir(105)); dot("$I$", I, dir(45)); [/asy][/asy] Let $I$ be the incenter, $X$ be the point where the incircle touches $\overline{BC}$, and $A'$ be the reflection of $A$ over $X$. I claim that $\overline{DP}$ passes through $A'$, which is fixed. Let $D'$ be the reflection of $D$ over $X$, and similarly define $E'$, $F'$. Note that, by construction, $\overline{BI}$, $\overline{CI}$ are the perpendicular bisectors of $\overline{DF}$, $\overline{DE}$ respectively. Thus $I$ is the circumcenter of $DEF$. It follows that $D'$ also lies on $(DEF)$ because $ID = ID'$. Similarly, we have $E', F' \in (DEF)$. Moreover, since $I$ is equidistant from the sides, we have $DD' = EE' = FF'$. The key claim is that $\overline{DP} \parallel \overline{AD'}$. Since we have $\overline{DF} \parallel \overline{D'F}$ and $\overline{DE} \parallel \overline{D'E}$, we need to check that \[\frac{\sin \angle F'D'A}{\sin \angle E'D'A} = \frac{\sin \angle D'F'A \cdot AF'}{\sin \angle D'E'A \cdot AE'} = \frac{D'F \cdot EF'}{D'E \cdot FE'}\]and \begin{align*} \frac{\sin \angle FDP}{\sin \angle EDP} &= \frac{\sin \angle DFP \cdot FP}{\sin \angle DEP \cdot EP} = \frac{\sin \angle DFC}{\sin \angle DEB} \cdot \frac{\sin \angle FEB}{\sin \angle EFC} \\ &= \frac{\sin \angle DFC}{\sin \angle EFC} \cdot \frac{\sin \angle FEB}{\sin \angle DEB} \\ &= \frac{\sin \angle FDD'}{\sin \angle FEE'} \cdot \frac{\sin \angle EFF'}{\sin \angle EDD'} \\ &= \frac{D'F \cdot EF'}{D'E \cdot FE'}. \end{align*}These are equal, so we have $\overline{DP} \parallel \overline{AD'}$. Finally, by construction $AD'A'D$ is a parallelogram, so $\overline{DA'} \parallel \overline{AD'}$. It follows that $D$, $P$, $A'$ are collinear as desired.

Let $XYZ$ be the intouch triangle of $ABC$ and $AB \cap XY = Q$. Since $AX, BY, CZ$ are concurrent at the Gergonne point of $ABC$, $(A, B; Z, Q) = -1$. Let $BC \cap EF = R$, $AP \cap BC = S$, and $AP \cap EF = T$. Since $AR, BE, CF$ are concurrent at $P$, $(B, C; S, R) = -1$. Also, $(T, R; F, E) = (E, F; T, R)^{-1} \overset{P}{=} (B, C; S, R)^{-1} = -1^{-1} = -1$. From $(A, B; Z, Q) = (T, R; F, E) = -1$, $XB \parallel DR$, $XZ \parallel DF$, and $XQ \parallel DE$, we have $XA \parallel DT$. Let $AX \cap PD = A'$. Then $(A, A'; X, \infty_{AX}) \overset{D}{=} (A, P; S, T) \overset{B}{=} (F, E; R, T) = (E, F; T, R) = -1$. Therefore, $A'$ is the reflection of $A$ with respect to $X$, which is a fixed point. Since $PD$ always passes through $A'$, we are done.