<Tri-2024-2143>, hint

Let PK extended cut  line BN in L', AK extended meet  BL' in N, <BNK=90. 

Since ABPK are concyclic< NKL'=1/2 <B, <NL'K=90-0.5<B=<BB(1)M, Line KB(1) cuts BC in M

at right angle.. Since <KL'B=<MB(1)B, BB(1)KL' are concyclic.

<KL'B(1)=KBB(1), <BKB(1)=<BL'B(1) and so KL'B=<MB(1)B.

Suppose Line B(1)A cuts line BN in L <BLB(1)=<BKB(1) and so. L=L'