Let $M$ be the midpoint of $BC$ and $E'$ be the reflection of $E$ with respect to $MI_a$. Also, let $\omega$ be the $A$-excircle. It is clear that $E'$ lies on $\omega$ and the problem is equivalent to proving that $DE'KE$ is cyclic. To do this we will show that $\angle E'DK = \angle E'EK$. Note that $EF \perp AI_a$ and $EE' \perp MI_a$. Therefore $\angle E'EK = \angle MI_aA$.

Assume that $X$ and $Y$ are the tangency points of $BC$ with the incircle and $\omega$, respectively. If $Y'$ is the reflection of $Y$ with respect to $I_a$ we can see that the dilation with center $A$ which sends the incircle to $\omega$, sends $X$ to $Y'$, hence $XY'$ passes through $A$.

Let $Z$ be the second intersection point of $XY'$ with $\omega$. We have $\angle YZX = 90^\circ$ and because $M$ is the midpoint of $XY$, we have $MZ = MY$. Therefore, $MZ$ is tangent to $\omega$. In fact $Z$ is the reflection of $Y$ with respect to $MI_a$. This implies that $MI_a \perp YZ$, hence $MI_a \parallel AZ$. Next, we claim that the triangles

$I_aZE'$ and $I_aAD$ are similar. Both of them are isosceles, so it is enough to show that $\angle ZI_aE' = \angle AI_aD$. Note that
$$\angle ZI_aE' = \angle YI_aE = \angle C, \quad \angle AI_aD = 2\angle AI_aB = \angle C$$

From the similarity of the triangles $I_aZE'$ and $I_aAD$, we can conclude that $I_aZA$ and $I_aE'D$ are similar and so
$$\angle E'DI_a = \angle ZAI_a = \angle MI_aA.$$
Therefore $DE'KE$ is cyclic as desired. ■