WLOG assume that $J$ is between $E,F.$ $\triangle JKL$ is obviously autopolar WRT the incircle $(I)$ of $\triangle ABC$ and $A \in LK,$ because its polar $EF$ WRT $(I)$ passes through $J.$ Analogously, $B \in LJ$ and $C \in JK.$ Since $KL$ does not cut $(I),$ then project $KL$ to infinity and $(I)$ into a circle, whose center is then $J.$ So $\triangle DEF$ becomes isosceles right at $D$ and $BCEF$ is a rectangle. Let $Y \equiv AM \cap BN \cap CP,$ $T \equiv KN \cap LP$ and $M' \equiv TJ \cap BC.$ Thus, it suffices to show that $M \equiv M'.$ Let $TP,TN$ cut $BC$ at $U,V.$ Then isosceles right $\triangle TUV$ and $\triangle JBC$ are homothetic through $M'$ $\Longrightarrow$ $\frac{\overline{M'B}}{\overline{M'C}}=\frac{\overline{M'U}}{\overline{M'V}}=\frac{\overline{BU}}{\overline{CV}}=\frac{\overline{PB}}{\overline{NC}} \Longrightarrow YM' \parallel PB \parallel NC \Longrightarrow M \equiv M'.$

Lemma:- In a triangle $\triangle ABC$, $P$ is a point inside it.Let $\triangle P_aP_bP_c$ be the cevian triangle of $P$.Consider a point $Q$ on the circumference of $\odot P_aP_bP_c$.Let $X=P_bP_c\cap P_aQ$.Analogously define $Y,Z$.Then $AX,BY,CZ$ are concurrent. Proof:- Let us consider a projective transformation that takes the circle $\odot P_aP_bP_c$ to the circle $P_a'P_b'P_c'$ and the tri-liner polar of $P$ to infinity(It can be proven that there exists such a projective transformation).Let $A'B'C'$ be the image of $ABC$.Clearly $P'$ becomes the centroid of $\triangle A'B'C'$.Now the result of the lemma is trivially true for the medial triangle.Hence the lemma is proven. Main Proof:- Now we just need to prove that poles of $JM,NK,LP$ wrt the incircle are collinear.Note that $J$ is the pole of $KL$.Similarly $K$ is the pole of $JL$ and $L$ is the pole of $JK$.Now clearly polars of $M,N,P$ pass through $D,E,F$ respectively.Since $AM,BN,CP$ are concurrent, clearly $\triangle A_1B_1C_1$ is perspective with $\triangle DEF$ where $\triangle A_1B_1C_1$ is the triangle formed by the polars of $M,N,P$.So $\triangle DEF$ is a cevian triangle of $\triangle A_1B_1C_1$.Now using the above lemma we conclude that $JM,NK,LP$ are concurrent.

My solution: Let $I$ be the incenter of $\triangle ABC$ . From Cevian nest theorem we get $AJ, BK, CL$ are concurrent at $Z$ . Since $KL$ is the polar of $J$ WRT $(I)$ , so $K, A, L$ are collinear . Similarly, we can prove $B \in LJ$ and $C \in JK$ . Since $\triangle ABC$ is the $Z$ -cevian triangle of $\triangle JKL$ , so from Cevian nest theorem we get $JM, KN, LP$ are concurrent . Q.E.D

What's cevian-nest theorem?

Dear Mathlinkers, see for example http://jl.ayme.pagesperso-orange.fr/ vol. 3 the cevian nests theorem or on search on mathlinks Sincerely Jean-Louis

cevian nest theorem, twice