Consider the lengths of the sides. How can we relate these to angles? In particular, the length $AB=\sqrt{3}+1$ seems useful.

Try experimenting with auxiliary points and lines.

Note: I highly suggest you try this out first. The solution will be very clear after this hint. Here's the hintPlot point $P$ on $AB$ so that $AP=1$ and $BP=\sqrt{3}$ [asy][asy] unitsize(2cm); pair A,B,C,D,M,K,P; A=(1,1.732); B=(3.732,1.732); C=(2.732,0); D=(0,0); M=(0.5,0.866); K=(2.116,1.299); P=(2,1.732); draw(A--B--C--D--cycle); draw(B--M); draw(C--K); draw(M--P,red); label("A",A,N); label("B",B,N); label("C",C,S); label("D",D,S); label("M",M,W); label("K",K,N); label("P",P,N); [/asy][/asy]

Consider an arbitary point $X$ on $AB$ such that $AX = AM=1$ then connect $XM$. Reflect $X$ across $AM$. Note that $\angle MAX = 120^{\circ}$ then angle $MAX' = 60^{\circ}$ Which implies triangle $MAX'$ is an equilateral triangle. Now also notice that triangle $MAX$ is an isosceles triangle and recall angle $MAX$ being $120^{\circ}$ then this implies $\angle AMX = 30, \angle AXM = 30$. From here we can say that $\angle X'MX = 90^{\circ}$ because $\angle X'MA + \angle AMX = 60^{\circ} + 30^{\circ} = 90^{\circ}$. Recall these facts from our diagram $X'M = 1, X'X = 2$ using Pythagorean theorem we have $YM = \sqrt{3}$. We knew that $AB = 1 + \sqrt{3}$ and $AX = 1$ which implies $XB = \sqrt{3}$. From there we can say that triangle $MXB$ is an isosceles triangle. Since $\angle MXB = 150^{\circ}$ hence $\angle XMB = \angle MBX = 15^{\circ}$ since $\angle ABC = 120^{\circ}$ then $\angle KBC = 105$. From the sum of interior angles we get $\angle CKB = 75^{\circ}$.