Solution:
Let $I$ be the incenter of $ABC$ and let $Z$ be the foot of the perpendicular from $K$ on $EC$. Since $KB$ is the bisector of $\hat{B}$, then $\angle EBC=15^{\circ}$ and since $EM$ is the perpendicular bisector of $BC$, then $\angle ECB=\angle EBC=15^{\circ}$. Therefore $\angle KEC=30^{\circ}$. Moreover, $\angle ECK=60^{\circ}-15^{\circ}=45^{\circ}$. This means that $KZC$ is isosceles and thus $Z$ is on the perpendicular bisector of $KC$.
Since $\angle KIC$ is the external angle of triangle $IBC$, and $I$ is the incenter of triangle $ABC$, then $\angle KIC=15^{\circ}+30^{\circ}=45^{\circ}$. Thus, $\angle KIC=\frac{\angle KZC}{2}$. Since also $Z$ is on the perpendicular bisector of $KC$, then $Z$ is the circumcenter of $IKC$. This means that $ZK=ZI=ZC$. Since also $\angle EKZ=60^{\circ}$, then the triangle $ZKI$ is equilateral. Moreover, since $\angle KEZ=30^{\circ}$, we have that $ZK=\frac{EK}{2}$, so $ZK=IK=IE$.
Therefore $DI$ is perpendicular to $EK$ and this means that $DIKA$ is cyclic. So $\angle KDI=\angle IAK=45^{\circ}$ and $\angle IKD=\angle IAD=45^{\circ}$. Thus $ID=IK=IE$ and so $KD$ is perpendicular to $DE$ as required.

Alternative Solution by PSC.
Let $P$ be the point of intersection of $EM$ with $AC$. The triangles $ABC$ and $MPC$ are equal since they have equal angles and $MC=\frac{BC}{2}=AC$. They also share the angle $\hat{C}$, so they must have identical incenter.
Let $I$ be the midpoint of $EK$. We have $\angle PEI=\angle BEM=75^{\circ}=\angle EKP$. So the triangle $PEK$ is isosceles and therefore $PI$ is a bisector of $\angle CPM$. So the incenter of $MPC$ belongs on $PI$. Since it shares the same incenter with $ABC$, then $I$ is the common incenter. We can now finish the proof as in the first solution.

Alternative Solution by PSC.
Let $P$ be the point of intersection of $EM$ with $AC$ and let $I$ be the midpoint of $EK$. Then the triangle $PBC$ is equilateral. We also have $\angle PEI=\angle BEM=75^{\circ}$ and $\angle PKE=75^{\circ}$, so $PEK$ is isosceles. We also have $PI \perp EK$ and $DI \perp EK$, so the points $P, D, I$ are collinear.
Furthermore, $\angle PBI=\angle BPI=45^{\circ}$, and therefore $BI=PI$.
We have $\angle DPA=\angle BEM=15^{\circ}$ and also $BM=\frac{AB}{2}=AC=PA$. So the right-angled triangles $PDA$ and $BEM$ are equal. Thus $PD=BE$.
So
$$EI=BI-BE=PI-PD=DI$$
Therefore $\angle DEI=\angle IDE=45^{\circ}$. Since $DE=DK$, we also have $\angle DEI=\angle DKI=\angle KDI=45^{\circ}$. So finally, $\angle EDK=90^{\circ}$.

Coordinate Geometry Solution by PSC.
We may assume that $A=(0,0), B=(0, \sqrt{3})$ and $C=(1,0)$. Since $m_{BC}=-\sqrt{3}$, then $m_{EM}=\frac{\sqrt{3}}{3}$. Since also $M=\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$, then the equation of $EM$ is $y=\frac{\sqrt{3}}{3} x+\frac{\sqrt{3}}{3}$. The slope of $BK$ is
$$m_{BK}=\tan \left(105^{\circ}\right)=\frac{\tan \left(60^{\circ}\right)+\tan \left(45^{\circ}\right)}{1-\tan \left(60^{\circ}\right) \tan \left(45^{\circ}\right)}=-(2+\sqrt{3})$$
So the equation of $BK$ is $y=-(2+\sqrt{3}) x+\sqrt{3}$ which gives $K=(2 \sqrt{3}-3,0)$ and $E=(2-\sqrt{3}, \sqrt{3}-1)$. Letting $I$ be the midpoint of $EK$ we get $I=\left(\frac{\sqrt{3}-1}{2}, \frac{\sqrt{3}-1}{2}\right)$. Thus $I$ is equidistant from the sides $AB, AC$, so $AI$ is the bisector of $\hat{A}$, and thus $I$ is the incenter of triangle $ABC$. We can now finish the proof as in the first solution.

Metric Solution by PSC.
We can assume that $AC=1$. Then $AB=\sqrt{3}$ and $BC=2$. So $BM=MC=1$. From triangle $BEM$ we get $BE=EC=\sec \left(15^{\circ}\right)$ and $EM=\tan \left(15^{\circ}\right)$. From triangle $BAK$ we get $BK=\sqrt{3} \sec \left(15^{\circ}\right)$. So $EK=BK-BE=(\sqrt{3}-1) \sec \left(15^{\circ}\right)$. Thus, if $N$ is the midpoint of $EK$, then $EN=NK=\frac{\sqrt{3}-1}{2} \sec \left(15^{\circ}\right)$ and $BN=BE+EN=\frac{\sqrt{3}+1}{2} \sec \left(15^{\circ}\right)$. From triangle $BDN$ we get $DN=BN \tan \left(15^{\circ}\right)=\frac{\sqrt{3}+1}{2} \tan \left(15^{\circ}\right) \sec \left(15^{\circ}\right)$. It is easy to check that $\tan \left(15^{\circ}\right)=2-\sqrt{3}$. Thus $DN=\frac{\sqrt{3}-1}{2} \sec \left(15^{\circ}\right)=EN$. So $DN=EN=EK$ and therefore $\angle EDN=\angle KDN=45^{\circ}$ and $\angle KDE=90^{\circ}$ as required.