Since $ A_1K = A_2K = B_1K = B_2K$, $ A_1A_2B_2B_1$ is cyclic. Call this circle $ \omega$. Since $ \angle O_1A_1K = \angle O_1B_1K = 90^\circ$, $ O_1$ is the pole of $ A_1B_1$ wrt. $ \omega$. By symmetry, $ O_2$ is the polar of $ A_2B_2$ wrt. $ \omega$. Hence $ A_1B_1\cap A_2B_2$ is the pole of $ O_1O_2$. An inversion wrt. $ \omega$ maps $ K$ to $ P$ so $ B_1K^2 = KP\cdot KL$. But then it follows from a SAS triangle similarity that $ \angle KB_1P = \angle KLB_1$. (1) Let $ KL\cap B_1B_2 = Z$. It is clear that $ A_1A_2$ is antiparallel to $ B_1B_2$ wrt. $ \angle A_1LA_2$. So since $ LK$ is a median of $ \triangle LA_1A_2$, $ LZ$ is a symmedian of $ \triangle LB_1B_2$. Also $ KB_1 = KB_2$. Using those two facts, it follows by a well-known lemma that $ KB_1$ and $ KB_2$ are the tangents to the circumcircle of $ \triangle KB_1B_2$. But then $ \angle B_2B_1K = \angle B_1LB_2$. (2) Subtracting (1) from (2) gives $ \angle PB_1L = \angle PLB_2$. But this implies that $ LB_2PB_1$ is cyclic, as desired. --- I wouldn't call this an 'easy' solution. It uses too many high powered theorems. What is your solution and/or the official one?

Alternatively it can be proven that $ KL\perp O_{1}O_{2}$ which is easily proven by the equivalence with $ LO_{2}^2 + KO_{1}^2 = LO_{1}^2 + KO_{2}^2$ wherei denote the center of the circles respectively by $ O_{1},O_{2}$.To prove this relation just notice that $ LA_{1}\perp KO_{2}$ and $ KO_{2} \perp LA_{2}$ so $ LO_{1}^2 = LK^{2} + A_{1}O_{1}^2 - A_{1}K^2$ and the similar one so the statement to prove reduces to $ KO_{2}^2 + A_{1}O_{1}^2 = KO_{1}^2 + A_{2}O_{2}^2$ which is trivial.Now form this we have $ KO_{2}BP$ cyclic so $ \angle BPL = \angle KO_{2}B = \angle LA_{1}K$ and to finish the proof notice that $ B_{1}$ and $ B_{2}$ are the feet of the altittudes in triangle $ LA_{1}A_{2}$.

My solution isn't nice. 1-step. In this step we will show that $ O_1O_2\perp LK$. It is well-known that $ A_2B_2\perp KO_2$.Assume that $ A_2B_2\cap O_1O_2 = T$ and $ A_2B_2\cap KO_2 = M$. Then $ \frac {PK}{KO_2} = \sin \angle PO_2K$ and $ \frac {O_2M}{MT} = \tg \angle MTO_2$.Hence $ \frac {PK}{KO_2}\cdot\frac {O_2M}{MT} = \frac {LM}{LK}$ (1). Since points $ P,T,O_2$ are collinear we have $ \frac {LP}{PK}\cdot \frac {KO_2}{O_2M}\cdot \frac {MT}{TL} = 1$ or $ \frac {PK}{KO_2}\cdot\frac {O_2M}{MT} = \frac {LP}{TL}$ (2). Comparing (1) and (2) we have that $ \frac {LM}{LK} = \frac {LP}{TL}$ or $ LM\cdot LT = LK\cdot LP$ (3). (3) means that $ K,P,T,M$ are concyclic, i.e $ \angle KPT = 90^\circ$. Since $ \angle KPO_2 = 90^\circ$ points $ K,P,O_2,A_2$ are also concyclic.But $ K,B_2,O_2,A_2$ are concyclic.Hence $ K,P,B_2,A_2$ are concyclic.Then $ \angle LPB_2 = \angle B_2A_2K$ (4). Since $ \angle LB_1B_2 = \angle B_2A_2K$ and from (4) we have $ \angle LPB_2 = \angle LB_1B_2$, i.e points $ B_1,L,B_2,P$ are concyclic.

Altheman wrote: I wouldn't call this an 'easy' solution. It uses too many high powered theorems. What is your solution and/or the official one? This is a solution,that i gave at olympiad: Since $ KA_1=KA_2=KB_1=KB_2$,then $ A_1B_1B_2A_2$ is the cyclic quadrilateral.Hence $ LA_1\cdot LB_1=LA_2\cdot LB_2$.Hence $ L$ lies on the radical axe of the circles $ \omega_1$ and $ \omega_2$,hence $ KL$ is a radical axe of these two circels.So $ KL\perp O_1O_2$.Now the problem became obvious,since $ O_1B_1PKA_1$ is cyclic,hence $ A_1B_1PK$ is a cyclic quadrilateral,hence $ \angle LPB_1=\angle B_1A_1A_2=\angle B_1B_2L$ and we are done. The official solution is longer and not so beautiful

My Solution Lemma With problem conditions $KL$ is the Radical Axis of the cicles $\omega_1$ and $\omega_2$. Proof Since $A_1K=A_2K=B_1K=B_2K$ that means $\angle A_1B_1A_2=\angle A_1B_2A_2=90^{\circ}$ that means the points $A_1,A_2,B_1$ and $B_2$ lie on the circle with diametr $A_1A_2$ and radius $A_1K=A_2K=B_1K=B_2K$. Then if we draw two tangents to the circles $LC_1$ and $LC_2$ then we have $|LC_1|^2=|LB_1|\cdot |LA_1|$ and $|LC_2|^2=|LB_2|\cdot |LA_2|$ We previously proved that $A_1,A_2,B_1$ and $B_2$ are cyclic and $|LB_1|\cdot |LA_1|=|LB_2|\cdot |LA_2|$. That means $|LC_1|=|LC_2|$ Which proves the Lemma. Since $KL$ is radical axis of $\omega_1$ and $\omega_2$ we have $KL\perp O_1O_2$. We also have $O_1B_1\perp KB_1$ and $O_2B_2\perp KB_2$ $\longrightarrow$ $O_1B_1PK$ and $O_2B_2PK$ are cyclic quadrilaterals. It follows that $\angle KB_1P=\angle KO_1P=\angle KA_1P$ ($A_1O_1PK$ is cyclic) , $\angle KB_2P=\angle KO_2P=\angle KA_2P$ ($A_2O_2PK$ is cyclic). Then $A_1B_1PK$ and $A_2B_2PK$ are cyclic quadrilaterals It follows that $\angle B_1A_1K=angle B_1B_2L=B_1PL$ Then $\boxed{ B_1,L,B_2,P \ \text{are cyclic}}$

Too easy for P5. Let $(KA_1O_1B_1)\cap (KA_2B_2O_2)=P'$. Since $A_1B_1A_2B_2$ is cyclic $LB_1\cdot LA_1=LB_2\cdot LA_2$, so $L-P'-K$ are collinear. Since $\angle O_1P'K=\angle O_1A_1K=90=\angle KA_2O_2=\angle O_2P'K$ we get $O_1-P'-O_2$ are collinear. So $P'=KL\cap O_1O_2$ so $P'\equiv P$. Miquel theorem on $\triangle LA_1A_2$ implies that $LB_1PB_2$ is cyclic. So done!