Let $P$ be the Miquel point of $ABCMLK$; then spiral similarity at $P$ takes $AL$ to $KC$, thus $PA = PK$ and $\angle{APK} = \angle{BAD} \implies P$ is the $B$-antipode wrt $(BAK)$ and thus also the $L$-antipode wrt $(ALM)$, which means it suffices to show that $C, P, N$ are collinear, AKA that $\angle{NAL} = 2 \cdot \angle{PLM}$, but this is obvious since $\angle{ABC} = 2 \cdot \angle{ABP}$. $\square$

$AK\cap DC=P$ $\angle KPC=\angle KAB=\angle DAK=\angle AKC \implies AL=KC=CP$ $\implies LAM \cong CPM \implies LM=MC$ $\angle NLM=\angle NAM=\angle MAB=\angle LNM \implies LM=MN$ $\implies LM=MN=MC$ $\implies \angle CNL=90$

First, $ALMN$ is concyclic and $AM$ bisects $\angle LAN$, so $LM=NM$. Next, $\angle BKA=\angle KAD=\angle BAK$ implies $AB=BK$, and Menelaus gives $$\frac{BA}{AL}\frac{LM}{MC}\frac{CK}{KB}=1,$$so $LM=MC$. Hence $M$ is the circumcentre of $\triangle LNC$ and lies on the side $LC$, so $\angle CNL=90^\circ$, as desired.