IMO Shortlist

We have $\angle APB = \angle ACB$ in the circumcircle and $\angle ACB = \angle A'XC$ because $A'X \parallel AC$. Hence, $\angle APB = \angle A'XC$, and so quadrilateral $BPA'X$ is cyclic. Similarly, it follows that $CYA'P$ is cyclic.
Now we are ready to transform $\angle KIL + \angle YPX$ to the sum of angles in triangle $A'CB$. By a homothety of factor 2 at $A$ we have $\angle KIL = \angle CA'B$. In circles $BPA'X$ and $CYA'P$ we have $\angle APX = \angle A'BC$ and $\angle YPA = \angle BCA'$, therefore
$$\angle KIL + \angle YPX = \angle CA'B + (\angle YPA + \angle APX) = \angle CA'B + \angle BCA' + \angle A'BC = 180^{\circ}.$$

Let $BC = a$, $AC = b$, $AB = c$ and $s = \frac{a+b+c}{2}$, and let the radii of the incircle, $B$-excircle and $C$-excircle be $r, r_b$ and $r_c$, respectively. Let the incircle be tangent to $AC$ and $AB$ at $B_0$ and $C_0$, respectively; let the $B$-excircle be tangent to $AC$ at $B_1$, and let the $C$-excircle be tangent to $AB$ at $C_1$. As is well-known, $AB_1 = s-c$ and $\text{area}(\triangle ABC) = rs = r_c(s-c)$.
Let the line through $X$, parallel to $AC$ be tangent to the incircle at $E$, and the line through $Y$, parallel to $AB$ be tangent to the incircle at $D$. Finally, let $AP$ meet $BB_1$ at $F$.

It is well-known that points $B, E$, and $B_1$ are collinear by the homothety between the incircle and the $B$-excircle, and $BE \parallel IK$ because $IK$ is a midline in triangle $B_0EB_1$. Similarly, it follows that $C, D$, and $C_1$ are collinear and $CD \parallel IL$. Hence, the problem reduces to proving $\angle YPA = \angle CBE$ (and its symmetric counterpart $\angle APX = \angle DCB$ with respect to the vertex $C$), so it suffices to prove that $FYPB$ is cyclic. Since $ACPB$ is cyclic, that is equivalent to $FY \parallel B_1C$ and $\frac{BF}{FB_1} = \frac{BY}{YC}$.
By the angle bisector theorem we have
$$\frac{BF}{FB_1} = \frac{AB}{AB_1} = \frac{c}{s-c}.$$
The homothety at $C$ that maps the incircle to the $C$-excircle sends $Y$ to $B$, so
$$\frac{BC}{YC} = \frac{r_c}{r} = \frac{s}{s-c}.$$
So,
$$\frac{BY}{YC} = \frac{BC}{YC} - 1 = \frac{s}{s-c} - 1 = \frac{c}{s-c} = \frac{BF}{FB_1},$$
which completes the solution.