Let $OP=l$ Let perpendiculars from $O$ to $AC$ and $BD$ be $OX$ and $OY$. and radius of circle be $r$. Also $K,L,M,N$ are on the same line $\perp$ to $OP$. $R_ {BOP} = \frac{lr}{2OY} $,$R_{COP} =\frac{lr}{2OX} $. $R_ {DOP} = \frac{lr}{2OY} $,$R_{AOP} =\frac{lr}{2OX} $. So $R_{BOP}=R_{DOP}$ and $ R_{COP} =R_{AOP}$ Easy to see $\angle KOL=\angle MON$ (by using $\angle OAP=\angle OCP$ Aand $\angle OBP =\angle ODP$ ) Hence $\triangle MON$ is congruent to $\triangle KOL$ Hence done.

Firstly, observe that $K,L,M,N$ are all co-linear on the perpendicular bisector of $OP$. Call the mid-pt. of $OP$ as $X$. Now, $\angle XPK= \frac{\pi}{2}-\angle PKX= \frac{\pi}{2}- \frac{1}{2}\angle PKO= \frac{\pi}{2}-\angle OAP$. Similarly, $\angle XPL= \frac{\pi}{2}- \angle OBP, \angle XPM= \frac{\pi}{2}- \angle OCP$ and $\angle XPN= \frac{\pi}{2}- \angle ODP$. So, $\sin \angle OAP= \cos \angle XPK = \frac{OP}{2R_A}$ and similarly, $\sin \angle OBP= \frac{OP}{2R_B}, \sin \angle OCP = \frac{OP}{2R_C}, \sin \angle ODP= \frac{OP}{2R_C}$. But, $ \angle OAP= \angle{OCP}, \angle OBP= \angle ODP \implies R_A=R_C,R_B=R_D$ $ \implies KO=MO, LO=NO \implies KX=MX, LX= NX \implies KL=MN$.

It is easy to prove that $K,L,M,N$ all lie on the straight line which perpendicularly bisects $OP$ As the line joining the centres of two intersecting circles perpendicularly bisects the common chord. Let the line joining the centres be $\ell$ Let the radius of $\odot AOP,\odot BOP,\odot COP,\odot DOP$ be $r_1,r_2,r_3,r_4$ respectively. Let $\ell\cap OP=X$ $LK=LX-KX=r_2\cos\angle PBO-r_1\cos\angle PAO$ By Sine rule, we have $\frac{OP}{\sin \angle OAP}=2r_1,\frac{OP}{\sin\angle OBP}=2r_2$ $\implies r_2\cos\angle PBO-r_1\cos\angle PAO$ $=\frac{OP}{2}\left(\frac{\cos\angle PBO}{\sin \angle PBO}-\frac{\cos \angle PAO}{\sin \angle PAO}\right)$ $=\frac{OP}{2}(\cot\angle PBO-\cot\angle PAO)$ Similarly, we can prove $MN=MX-XM=\cot\angle PCO+\cot\angle PBO$ As $\angle PAO=\angle PCO$ and $\angle PBO=\angle PDO$ We get $KL=MN$.

$\angle ADC= \alpha,\angle DCB = \beta$ $OP = x$ $R_{AOP} = \frac{x}{2cos \alpha} , R_{OPB}= \frac{x}{2cos \beta}$ $$KL^2 = \frac{x^2}{(2cos \alpha)^2} + \frac{x^2}{(2cos \beta)^2} - \frac{x^2 \cdot cos(\alpha + \beta)}{2cos \alpha \cdot cos \beta}$$This is symmetric for both $180 - \alpha$ and $180 -\beta$. $\blacksquare$