Solution

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The hardest part is drawing the diagram. [asy][asy] unitsize(100); pen n_purple = rgb(0.7,0.4,1), n_blue = rgb(0,0.6,1), n_green = rgb(0,0.4,0), n_orange = rgb(1,0.4,0.1), n_red = rgb(1,0.2,0.4); pair A, B, C, P, E, F, Bp, Cp, Bpp, Cpp, X; path w = unitcircle; A = dir(140); B = dir(210); C = dir(330); //P = (-0.2, -0.3); P = (-0.4, -0.1); E = extension(B, P, C, A); F = extension(C, P, A, B); Bp = point(w, intersections(w, E, F)[0]); Cp = point(w, intersections(w, E, F)[1]); Bpp = extension(Cp, P, B, C); Cpp = extension(Bp, P, B, C); X = extension(Bp, Bpp, Cp, Cpp); draw(B--Bpp^^C--Cpp, gray(0.6)); draw(Bp--Cpp^^Cp--Bpp, gray(0.6)); draw(Bp--Cp, gray(0.6)); draw(A--B--C--cycle); draw(B--E^^C--F); draw(Bpp--X^^Cpp--X, n_orange + dashed); draw(w, n_blue); dot(A^^B^^C^^P^^E^^F^^Bp^^Cp^^Bpp^^Cpp^^X); label("$A$", A, dir(140)); label("$B$", B, dir(240)); label("$C$", C, dir(300)); label("$P$", P, dir(280)); label("$E$", E, dir(80)); label("$F$", F, dir(140)); label("$B'$", Bp, dir(40)); label("$C'$", Cp, dir(160)); label("$B''$", Bpp, dir(270)); label("$C''$", Cpp, dir(270)); [/asy][/asy] Let $\omega$ be the circumcircle of $\triangle ABC$. Then \[(BC; B''C'') \stackrel{P}{=} (EF; C'B') \stackrel{A}{=} (CB; C'B')_{\omega} = (BC; B'C')_{\omega}\]implying lines $B'B''$ and $C'C''$ meet on $\omega$.

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My solution is similar to cantonmathguy solution. Let Q is the point when C'C" meet circumcircle.QB' meet BC at B"' For point P we have (C'FEB')=(C"BCB") In the circle we have (QC',QB,QC,QB')=(C"BCB"')=(AC',AB,AC,AB')=(C'FEB') So (C"BCB"')=(C"BCB") So B"' is point B" and we are done

it can be dealt much easier, the problem is purely projective so we can transfer $P$ anything then we shall select the centroid of triangle. as we put $P$ the centroid, just giving hints to solve $\triangle B''CP \sim \triangle PFC'$ implies $B''C=2FC'$ and similarly $BC''=2B'E \implies 2B'C'=B''C''$ and because they're parallel then we see $B'$ is the midpoint of $KB''$ and $C'$ is the midpoint of $KC''$ so $P \equiv G_{KB''C''}$ and with the fact provided the rest is straightforward (you can even deal with complex numbers)

SDEIBAR wrote: My solution is similar to cantonmathguy solution. Let Q is the point when C'C" meet circumcircle.QB' meet BC at B"' For point P we have (C'FEB')=(C"BCB") In the circle we have (QC',QB,QC,QB')=(C"BCB"')=(AC',AB,AC,AB')=(C'FEB') So (C"BCB"')=(C"BCB") So B"' is point B" and we are done beautiful

a different solution : Let $D$ be intersection of $AB',BC$ and $K$ be intersection of $EF,BC$ and $R$ be second intersection of $AK$ and circumcircle of $ABC$ We will show that intersection of $B'B''$, $C'C''$ is $R$ if intersection of $RC', BC$ be $C_1$ is enough to show $B'C_1, EC,FB$ are concurrent or: $(B'E, FK)=(C_1C, BK) $ $(B'E, FK) \stackrel{A} {=} (DB, CK) = (KC, BD) \stackrel{B'} {=} (C'C, BA) \stackrel{R} {=} (C_1C, BK)$ and we are done.

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