Extend the diagram such that lines $MK$ and $NK$ intersect the circle at $E$ and $F$, respectively. Hence, by simple angle chase, where first equality comes from tangent-chord theorem and third equality comes from the inscribed angle theorem; $$\measuredangle MKA =\measuredangle MNK =\measuredangle MNF =\measuredangle MEF,$$thus $AB\parallel EF.$ Thus $\measuredangle KNB=\measuredangle FNB= \measuredangle FAB= \measuredangle ABE =\measuredangle AME=\measuredangle AMK$, where second one and forth one comes from the inscribed angle theorem and the third one comes from knowing that $ABEF$ is an isosceles trapezium. Hence, we are done.