I think the answer is $\frac{\sqrt{3}+1}{2}$.

We use complex numbers. Since $AQB$, $BRC$, $CSD$ and $DPA$ are equilateral, we have \begin{align*} q &=e^{i\pi/3}(b-a)+a \\ r &=e^{i\pi/3}(c-b)+b \\ s &=e^{i\pi/3}(d-c)+c \\ p &=e^{i\pi/3}(a-d)+d. \end{align*} Let $\omega=e^{i\pi /3}$. Thus we have \begin{align*} KM=|k-m| &=\left| \dfrac{p+q}{2}-\dfrac{r+s}{2}\right| \\ &=\left| \omega(b-d)+\dfrac{a+d-b-c}{2}\right| \\ &=\left| \dfrac{1}{2}(a-c)+\left(\omega-\dfrac{1}{2}\right)(b-d)\right|. \end{align*}Note that $LN$ has a similar expression. Therefore we have \begin{align*} \dfrac{KM+LN}{AC+BD} &=\dfrac{\left| \dfrac{1}{2}(a-c)+\left(\omega-\dfrac{1}{2}\right)(b-d)\right| +\left| \left(\omega-\dfrac{1}{2}\right)(a-c)+\dfrac{1}{2}(b-d)\right|}{|a-c|+|b-d|} \\ &\le \dfrac{\left(\left| \omega-\dfrac{1}{2}\right|+\dfrac{1}{2}\right)|a-c|+\left(\left| \omega-\dfrac{1}{2}\right|+\dfrac{1}{2}\right)|b-d|}{|a-c|+|b-d|} \\ &=\left |\omega-\dfrac{1}{2}\right|+\dfrac{1}{2}=\dfrac{\sqrt{3}+1}{2}. \end{align*}