By the sine law on triangle $AB'P$,
$$\frac{AB'}{\sin \angle APB'} = \frac{AP}{\sin \angle AB'P},$$
so
$$AB' = AP \cdot \frac{\sin \angle APB'}{\sin \angle AB'P}.$$
[asy] import graph; import geometry; unitsize(0.5 cm); pair[] A, B, C; pair P, R; A[0] = (2,12); B[0] = (0,0); C[0] = (14,0); P = (4,5); R = 5*dir(70); A[1] = extension(B[0],C[0],P,reflect(P + R,P - R)*(A[0])); B[1] = extension(C[0],A[0],P,reflect(P + R,P - R)*(B[0])); C[1] = extension(A[0],B[0],P,reflect(P + R,P - R)*(C[0])); draw((P - R)--(P + R),red); draw(A[1]--B[1]--C[1]--cycle,blue); draw(A[0]--B[0]--C[0]--cycle); draw(A[0]--P); draw(B[0]--P); draw(C[0]--P); draw(P--A[1]); draw(P--B[1]); draw(P--C[1]); draw(A[1]--B[0]); draw(A[1]--B[0]); label("$A$", A[0], N); label("$B$", B[0], S); label("$C$", C[0], SE); dot("$A'$", A[1], SW); dot("$B'$", B[1], NE); dot("$C'$", C[1], W); dot("$P$", P, SE); label("$\gamma$", P + R, N); [/asy]
Similarly,
\begin{align*} B'C &= CP \cdot \frac{\sin \angle CPB'}{\sin \angle CB'P}, \\ CA' &= CP \cdot \frac{\sin \angle CPA'}{\sin \angle CA'P}, \\ A'B &= BP \cdot \frac{\sin \angle BPA'}{\sin \angle BA'P}, \\ BC' &= BP \cdot \frac{\sin \angle BPC'}{\sin \angle BC'P}, \\ C'A &= AP \cdot \frac{\sin \angle APC'}{\sin \angle AC'P}. \end{align*}
Hence,
\begin{align*} &\frac{AB'}{B'C} \cdot \frac{CA'}{A'B} \cdot \frac{BC'}{C'A} \\ &= \frac{\sin \angle APB'}{\sin \angle AB'P} \cdot \frac{\sin \angle CB'P}{\sin \angle CPB'} \cdot \frac{\sin \angle CPA'}{\sin \angle CA'P} \cdot \frac{\sin \angle BA'P}{\sin \angle BPA'} \cdot \frac{\sin \angle BPC'}{\sin \angle BC'P} \cdot \frac{\sin \angle AC'P}{\sin \angle APC'}. \end{align*}
Since angles $\angle AB'P$ and $\angle CB'P$ are supplementary or equal, depending on the position of $B'$ on $AC$,
$$\sin \angle AB'P = \sin \angle CB'P.$$
Similarly,
\begin{align*} \sin \angle CA'P &= \sin \angle BA'P, \\ \sin \angle BC'P &= \sin \angle AC'P. \end{align*}
By the reflective property, $\angle APB'$ and $\angle BPA'$ are supplementary or equal, so
$$\sin \angle APB' = \sin \angle BPA'.$$
Similarly,
\begin{align*} \sin \angle CPA' &= \sin \angle APC', \\ \sin \angle BPC' &= \sin \angle CPB'. \end{align*}
Therefore,
$$\frac{AB'}{B'C} \cdot \frac{CA'}{A'B} \cdot \frac{BC'}{C'A} = 1,$$
so by Menelaus's theorem, $A'$, $B'$, and $C'$ are collinear.