It is clear that $ABCD$ is a deltoid, thus $BD \perp AC$; that also means that $CA$ is a bisector of $\angle BCD$. Denote the incenter of $\triangle BCD$ as $I$, the intersection point of $AC$ and $BD$ as $X$ and $\angle BCA = 2\alpha$. $\angle CBX = 90^\circ - 2\alpha$ because $\triangle BXC$ is right-angled. Since $\triangle MBL$ is isosceles, $BI$ is an altitude to the $ML$ side and bisector of $\angle MBL$. That means that $\angle BLM = 90^\circ - \frac 1 2 \angle CBX = 90^\circ - (45^\circ - \alpha) = 45^\circ + \alpha$. Besides, $\angle BIX = \angle BLM$ since $\triangle BXI$ also is right-angled and $\angle XBI = 45^\circ - \alpha$. That means that points $B$, $L$, $I$ and $A$ are concyclic $(1)$. Since $\triangle XBI = \triangle XDI$ (as one is a mirror image of another), $\angle LIC = \angle XID = \angle XIB = 45^\circ + \alpha$. Now from $(1)$ it follows that $\angle BAI = \angle BIL + \angle IBL = (180^\circ - \angle BIX - \angle LIC) + \angle IBL = (180^\circ - 45^\circ - \alpha - 45^\circ - \alpha) + 45^\circ - \alpha = 135^\circ - 3\alpha$. Finally as $\angle BAD = 2\angle BAC = 2\angle BAI = 2\cdot (135^\circ - 3\alpha) = 270^\circ - 6\alpha$ and $\angle BCD = 2\angle BCA = 4\alpha$, $2\angle BAD + 3\angle BCD = 2\cdot (270^\circ - 6\alpha) + 3\cdot (4 \alpha) = 540^\circ - 12\alpha + 12\alpha = 540^\circ$, and that is the answer.