Let $\omega$ be the object circle and $\tau$ the perpendicular line to $AC$ through $C.$ $\tau$ is the inverse of $\omega$ under the inversion with pole $A$ and power $\overline{AK }\cdot \overline{AC}$ $\Longrightarrow$ $M' \equiv AB \cap \tau$ and $N' \equiv AD \cap \tau$ are the inverses of $M,N.$ Thus $\overline{AK} \cdot \overline{AC}=\overline{AM} \cdot \overline{AM'} =\overline{AN} \cdot \overline{AN'}.$ But from $\triangle M'BC \sim \triangle M'AN' \sim \triangle CDN',$ we deduce that $\frac{\overline{AB}}{\overline{AM'}}+\frac{\overline{AD}}{\overline{AN'}}=1 \Longrightarrow \overline{AB} \cdot \frac{\overline{AM}}{ \overline{AK} \cdot \overline{AC}}+\overline{AD} \cdot \frac{\overline{AN}}{\overline{AK} \cdot \overline{AC}}=1$ $\Longrightarrow \overline{AB} \cdot \overline{AM}+\overline{AD} \cdot \overline{AN}=\overline{AK} \cdot \overline{AC}.$

By Ptolemy: $AM*NK+AN*KM = AK*NM$ (1) Also $\angle{KMN}=\angle{KAN}=\angle{ACB}$ and $\angle{MNK}=\angle{MAK}$ so $ACB$ is similar to $NMK$ and therefore: $\frac{AC}{MN}=\frac{AB}{NK}=\frac{BC}{MK}=\frac{AD}{MK}= k$ which implies, $NK= AB*\frac{1}{k}$ , $ KM = AD*\frac{1}{k}$ and $NM= AC*\frac{1}{k}$ . Substituting this into (1) we get the desired result.

Place the origin at $A$. It is well known that the function $\pi(X) = \operatorname{Pow}_A(X) - \operatorname{Pow}_{(KMN)}(X)$ is linear. Consequently, \begin{align*} AC\cdot AK &= CA^2 - CA\cdot CK \\ &= \pi(C) \\ &= \pi(B+D) \\ &= \pi(B) + \pi(D) \\ &= AB\cdot AM + AD\cdot AN. \end{align*}

If the circle $\odot(MBK)$ intersects $AC$ second time at $P$ and circle $\odot(DNK)$ intersects $AC$ second time at $Q$, it's not big deal to prove $BP\parallel DQ$, wherefrom $PC=AQ\ (\ 1 \ )$ and applying p.o.p. we get $AN\cdot AD=AK\cdot AQ\ (\ 2\ )$ and $ AM\cdot AB=AK\cdot AP\ (\ 3\ )$. Adding $(2)$ and $(3)$ side by side and taking into account $(1)$, we get the desired relation. Best regards, sunken rock

Claim: $\Delta MKN \sim \Delta CBA$ Proof: Angle chase. Now from Ptolemy's theorem, $$AM \cdot NK + AN \cdot MK = AK \cdot MN$. Finding $MN, NK, MK$ wrt $AB,BC,CA$ and substituting gives the result. $\square$