This is unequivocally the best solution for this problem. Well, not your typical config geo, that's for sure!

Solution from Twitch Solves ISL: Let $E$ be the reflection of $A$ across $\overline{BC}$. Let $I$ and $J$ denote the incenters of $\triangle EBC$ and $\triangle DBC$. Redefine $Z = \overline{DE} \cap \overline{IPJ}$; then the problem is to show $X$, $Y$, $Z$ are collinear. [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ pair A = (10.00216,0.58182); pair B = (9.,-3.); pair C = (45.,-3.); pair P = (11.26936,-3.); pair J = (11.26936,-2.03424); pair D = (10.88129,-1.04465); pair Y = (10.00216,3.38498); pair Z = (11.26936,1.39956); pair K = (10.00216,-3.); pair E = (10.00216,-6.58182); pair I = (11.26936,-4.72156); pair W = (10.78721,-1.63719); size(16cm); pen zzttqq = rgb(0.6,0.2,0.); pen yqqqqq = rgb(0.50196,0.,0.); pen qqffff = rgb(0.,1.,1.); pen xfqqff = rgb(0.49803,0.,1.); pen qqwuqq = rgb(0.,0.39215,0.); pen cqcqcq = rgb(0.75294,0.75294,0.75294); draw(A--B--C--cycle, linewidth(0.6) + zzttqq); draw(B--D--C--cycle, linewidth(0.6) + qqffff); draw(B--E--C--cycle, linewidth(0.6) + zzttqq); draw(A--B, linewidth(0.6) + zzttqq); draw(B--C, linewidth(0.6) + zzttqq); draw(C--A, linewidth(0.6) + zzttqq); draw(circle((11.26936,-1.27843), 1.72156), linewidth(0.6) + yqqqqq); draw(circle(J, 0.96575), linewidth(0.6) + blue); draw(B--D, linewidth(0.6) + qqffff); draw(D--C, linewidth(0.6) + qqffff); draw(C--B, linewidth(0.6) + qqffff); draw(P--Y, linewidth(0.6) + xfqqff); draw(circle(I, 1.72156), linewidth(0.6) + yqqqqq); draw(B--E, linewidth(0.6) + zzttqq); draw(E--C, linewidth(0.6) + zzttqq); draw(C--B, linewidth(0.6) + zzttqq); draw(E--Z, linewidth(0.6) + qqwuqq); draw(I--Z, linewidth(0.6) + yqqqqq); draw(A--E, linewidth(0.6)); draw(A--P, linewidth(0.6)); draw(A--Y, linewidth(0.6)); dot("$A$", A, dir(160)); dot("$B$", B, dir(180)); dot("$C$", C, dir((-3794.550, 716.315))); dot("$P$", P, dir((4.522, 8.332))); dot("$J$", J, dir(0)); dot("$D$", D, dir(150)); dot("$Y$", Y, dir((4.454, 7.830))); dot("$Z$", Z, dir((4.522, 8.583))); dot("$E$", E, dir(270)); dot("$I$", I, dir(-45)); dot("$W$", W, dir(200)); clip(box( (8,-10), (15,10) ) ); [/asy][/asy] To interpret the condition about $P$: Claim: $(IJ;PZ) = -1$. Proof. Because the condition on $P$ forces $BD+CE=BE+CD$, there is an incircle for $BDCE$. Hence by Monge theorem, $Z$ coincides with the exsimilicenter of $(I)$ and $(J)$. As $I$ is the insimilicenter, the conclusion follows. $\blacksquare$ Next, let $W = \overline{ZDE} \cap \overline{AP}$. Claim: Triangles $DJP$ and $AWE$ are perspective, i.e.\ the lines $\overline{DA}$, $\overline{JW}$, $\overline{PE}$ are concurrent. Proof. We have $-1 = (ZP;JI) \overset{W}{=} (E,A;\overline{JW}\cap\overline{AE},Y)$. Then it follows by looking at complete quadrilateral $ADPEWY$. $\blacksquare$ Now by Desargue's theorem, it follows $\overline{DJ} \cap \overline{AW} = X$, $\overline{DP} \cap \overline{AE} = Y$, $\overline{JP} \cap \overline{WE} = Z$ are collinear.

As a note, another problem with a solution similar in flavour is ISL 2022/G6.

Easy there is a hyperbola with foci $B,C$ through $A,D$ and reflection $A'$ of $A$ over $BC$. Then Pascal's on $DDA'APP$ gives the desired