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Let me denote the circles with the diameters $AC$ and $BD$ as $\Gamma _1$ and $\Gamma _2$ respectively.
As, $O$ lies on the radical axis of $\Gamma_1$ and $\Gamma_2$, so,
$BO.NO=CO.OM$ which implies that $BCNM$ is a cyclic quadrangle.
Let, $AM\cap DN = K$ and $AD\cap XY = G$ .
As, $\angle ONK=\angle OMK=90^{\circ}$.
$OMKN$ is cyclic as well.
$OK$ is the diameter.

Now, We need to prove that, $K,O,G$ are collinear.
As, $\Delta MKO \sim \Delta OCG$,
$\frac{MO}{GO}=\frac{KO}{CO}=\frac{MK}{CG}$
$\Rightarrow \frac{CG.MO}{MK.CO}=\frac{GO}{CO}$.

Now, $\Delta AGM \sim \Delta AKD$ as $\angle AGM= \angle AKD$.(As $MKDG$ is cyclic)
So, $\frac{AG}{AK}=\frac{AM}{AC}$.

But, $\Delta AMC \sim \Delta COG$.
So,$\frac{CO}{GO}=\frac{AC}{AM}$.

Therefore,
$\frac{AK}{AG}=\frac{CO}{GO}$.

So,
$\frac{AK.MO.CG}{AG.MK.CO}=1$.
Using the converse of Menelaus's theorem for $\Delta AMC$, we conclude that, $K,O,G$ are collinear.
[Proved]

I have denoted the point $O$ as $P$ here. Sorry for the inconvinience.

Let $T$ be the intersection of $AM$ and $XY$(extended). Now $TQ$ is the perpendicualr to $AD$.$Q$ is the foot of perpendicular from $T$ to $AD$. Now, $\angle TMC=\angle TQC=90^{\circ}.$ So, the pointd $T,M,Q,C$ are concyclic. so $\angle MTQ=\angle MCQ=\angle MCB$.
Using power of point:
$PM.PC=PX.PY=BP.BN$. Therefore, $B,C,M,N$ are concylic. Now, $\angle BCM=\angle BNM , so \angle MTP=\angle MTQ=\angle BCM= \angle BNM=\angle PNM$. Therefore, points $M,T,P,N$ are concyclic. So, $\angle PNT= 180^{\circ}-\angle PMT=90^{\circ}$. So $\angle DNP+\angle PNT=90^{\circ}+90^{\circ}=180^{\circ}$. So, $D,N,T$ are collinear, Therefore $T$ is also the point of intersection of $DN$ and $XY$.