Too bad this problem did not get more attention. A lot of things going on here...
First, let's prove that $\triangle BB'C\cong \triangle BLC$
$BC'HA'$ is a cyclic quadrilateral so $\angle BA'C'=\angle BHC'=\angle CA'L=\alpha $.
$A'HB'C$ is also a cyclic quadrilateral so $\angle B'A'C=\angle B'HC=\alpha$.
So $\angle B'A'C=\angle LA'C$ which also means that $\angle BA'B'=\angle BA'L$.
Consider triangles $\triangle BA'B'$ and $\triangle BA'L$. They have one common side ($BA'$), a pair of equal sides ($BB'=BL$) and a pair of equal angles ($\angle BA'B'=\angle BA'L$).
$$\triangle BA'B'\cong \triangle BA'L$$
...which also means that $B'A'=A'L$. Now it's trivial to prove that:
$$\triangle B'A'C\cong \triangle LA'C$$
$$\triangle BB'C\cong \triangle BLC$$
Also note that in $\triangle A'CL$:
$$\angle A'CL=\angle A'CB'=\gamma, \space\angle CA'L=\alpha\implies A'LC=\beta$$
You can show in exactly the same way that:
$$\angle BKP=90^\circ, \space \angle C'KA=\beta$$
So triangles $\triangle BKL$ and $\triangle KLP$ are isosceles and quadrilateral $BKPL$ is symmetric with respect to $BP$.
Now let us prove that $\angle ABP=\angle B'BC=90^\circ-\gamma$
In triangle $\triangle ABP:$
$$\angle BAP=180^\circ-\angle BAK=180^\circ-\angle BAC=180^\circ-\alpha$$
$$\angle ABP=180^\circ-\angle BAP - \angle BPA=180^\circ-(180^\circ-\alpha)-(90^\circ-\beta)=90^\circ-\gamma$$
So we proved that:
$$\angle ABP=\angle B'BC$$
It's a well known (and easily provable) fact that bisector of $\angle B$ splits not only the angle itslef but also the angle between the height $BB'$ and the line that contains point $B$ and the center of circumscribed circle $O$ (for more details you can check this page or Wikipedia; if you still want the proof of this fact, please let me know). In other words, lines $BB'$ and $BO$ are symmetric with respect to bisector of $\angle B$.
In this particular case, $\angle ABP=\angle B'BC$ which means that lines $BB'$ and $BP$ are actually symmetric with respect to the bisector of $\angle B$ (not shown in the picture). This means that the line $BP$ must contain the center of circumscribed circle $O$. Done!
EDIT: Mick mentioned that I was using ASS to prove that triangles $\triangle BA'B'$ and $\triangle BA'L$ were congruent. I think that I had the right to do because in acute triangles the angle at $A'$ is always obtuse. Quote from Wikipedia: If two triangles satisfy the SSA condition and the length of the side opposite the angle is greater than or equal to the length of the adjacent side (SSA, or long side-short side-angle), then the two triangles are congruent. But, OK, I admit that ASS is ugly so let's prove the same without it. Maybe not the easiest proof but it still works:
So we want to prove (without ASS) that $A'B'=A'L$ knowing that:
$$BB'=BL, \space \angle BA'B'=\angle BA'L>90^\circ $$
Draw circumscribed circles for triangles $\triangle BA'B'$ and $\triangle BA'L$. Central angles $\angle BO_1L$ and $\angle BO_2B'$ are the same because they correspond to the same inscribed angle ($\angle BA'B'=\angle BA'L$).
It's now obvious that triangles $\triangle BO_1L$ and $\triangle BO_2B'$ have the same angles. Having in mind that $BB'=BL$ and by applying ASA:
$$\triangle BO_1L\cong\triangle BO_2B'$$
In other words, both circles have the same radius and, consequentially:
$$\angle BO_1L=\angle BO_2B',\space\angle BO_1A'=\angle BO_2A'$$
$$\implies \angle A'O_1L=\angle A'O_2B'\implies A'B'=A'L$$