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I am looking for a proof of a problem as follows:

Let ABC and $A'B'C'$ be two triangles circumcribed a conic. Let $M=AC \cap A'C'$, $N=AB \cap A'B'$. Let $a_{b'}=BC\cap A'C'$, $a'_c=B'C'\cap AB$, $a_{c'}=BC\cap A'B'$, $a'_b=B'C'\cap AC$. $P= a_{b'}a'_c \cap a_{c'}a'_b$. Then show that M, N, P are collinear

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    In other words Show that *Pairwise opposite sides in an ordered conic_ inscribed_hexagon cut along the same straight line* Right ? – 2017-01-12
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    $A, A', P, a_{b'}, a'_c , a_{c'}, a'_b$ lie on a conic – 2017-01-12

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Let $l = MN$. Then by Pascal's theorem for the inscribed in a conic hexagon $CABC'A'B'$ $$Q = BC' \cap B'C \, \, \in \,\, l$$ Let $MB \cap B'C' =\tilde{A}'$ and $NC' \cap BC =\tilde{A}$. Then, by Pappus' theorem for hexagon $a'_{c}B\tilde{A}'\tilde{A}C'a_{b'}$ with vertices on the pair of lines $BC$ and $B'C'$, the three points $M, \, N,$ and $P'=a_{b'}a'_c \cap \tilde{A}\tilde{A}'$ are collinear, i.e. $P' \in l = MN$.

Let $R = \tilde{A}C' \cap \tilde{A}'B$ and $S = Ca'_b \cap a_{c'}B'$. Next, apply the converse direction of Desargue's theorem to the pair of triangles $BRC'$ and $CSB'$, where $l = MN$ is their common perspective line with the three points $M,\, N, \, Q$ lying on $l$, and conclude that the three lines $BC, \, B'C'$ and $RS$ are concurrent.

Finally, apply Desargue's theorem to the pair of triangles $\tilde{A}R\tilde{A}'$ and $a_{c'}Sa'_b$, using the already established fact that the lines $BC, \, B'C'$ and $RS$ are concurrent. Thus the tree points $M,\, N$ and $P'' = \tilde{A}\tilde{A}' \cap a_{c'}a'_b$ are collinear, which means that $P'' \in l$. However, as already shown earlier, $l \cap \tilde{A}\tilde{A}' = P'$, which implies that $P'' = P'$ lies on the line $l=MN$. Since $$P = a_{b'}a'_c \cap a_{c'}a'_b = a_{b'}a'_c \cap l \cap \tilde{A}\tilde{A'} \cap a_{c'}a'_b = P'=P''$$ we conclude that point $P$ lies on the line $l$ determine by the points $M$ and $N$.

Another proof is as follows: let $S = Ca'_b \cap a_{c'}B'$ and $T = Ba'_c \cap C'a_{b'}$. Then by Pascal's theorem for the inscribed in a conic hexagon $CABC'A'B'$ $$BC' \cap B'C \, \, \in \,\, MN$$ But then, by Desargue's theorem for triangles $BTC'$ and $B'SC$ the three lines $BB', \, CC', \, ST$ are concurrent, which by Brianchon's theorem implies that the hexagon $BCSB'C'T$ is circumscribed around a conic. Consequently, the hexagon $Sa_{c'}a_{b'}Ta'_ca'_b$ is circumscribed around the same conic, so by Birachon's theorem the three lines $a_{b'}a'_b, \, a'_ca_{c'}, \, ST$ are concurrent. Finally, by Desargue's theorem for triangles $a_{b'}Ta'_c$ and $a'_bSa_{c'}$ the three intersection points $M, \, N$ and $P = a_{b'}a'_c \cap a_{c'}a'_b$ are collinear.

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This is Pascal's theorem (aka the Hexagrammum Mysticum Theorem) which holds good even when opposite sides of conic are extended. Standard proof e.g., in Wiki using cross-ratios.

Pascal's theorem

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    I don't Think this problem exactly the Pascal's theorem – 2017-01-12
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    In what way does it differ? – 2017-01-12
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    Can you using the Pascal theorem to proof the problem? – 2017-01-12
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    There is no need to freshly prove. The Wiki link specifically states " extended if necessary" and the second figure in the Wiki is your diagram. – 2017-01-12
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    Do You mean this link: https://en.m.wikipedia.org/wiki/Pascal's_theorem. And these word: The intersections of the extended opposite sides of simple cyclic hexagon ABCDEF (right) lie on the Pascal line MNP (left). – 2017-01-12
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    Please checked detail again. Because the Pascal theorem: let $P'=BC' \cap B'C$ then $M, N, P'$ are collinear. But this problem is $M, N, P$ are collinear. – 2017-01-12
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    Where is $P^{\prime}$ in your diagram or Wiki ? In any case it does not matter. – 2017-01-12
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    Define $P'$ in the comment above, In my diagram. – 2017-01-12
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    I may not deserve upvoting as at start I did not understand the finer point OP was making. – 2017-01-13
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I think $ABC$ and $A'B'C'$ are inscribed the conic. In this case by Poncelet's theorem for triangles, there is a conic inscribed their sides. Just use Brianchon's theorem for this conic: we get that $MN$, $a'_c a_{b'}$ and $a'_b a_{c'}$ are concurrent.

Another possible solution, without using Poncelet's theorem: the common point of$MN$, $a'_c a_{b'}$ and $a'_b a_{c'}$ is just the Kirkman-point $K(AB'BC'CA')$.