Let $I_{XYZ}$ be the incenter of triangle $XYZ$ and $I_{X,YZ}$ be the excenter of triangle $XYZ$ with respect to vertex $X$.We show that $I_{BCD}I_{ACD}I_{ABD}I_{ABC}$ is a rectangle. For this we show that $\angle {I_{ACD}I_{ABD}I_{ABC}}=90$.\[ \angle {AI_{ABD}I_{ACD}}=180-\angle{ADI_{ADC}}=180-\dfrac{\angle{ADC}}{2} \]\[ \angle{AI_{ABD}I_{ABC}}=180-\angle{ABI_{ABC}}=180-\dfrac{\angle{ABC}}{2} \]\[ \Rightarrow \angle{I_{ABC}I_{ABD}I_{ACD}}=90 \] So the assertion is true. Now we show that these points are collinear: $l_1: I_{C,AD},I_{ABD},I_{ABC},I_{D,BC}$ $l_2: I_{B,AD},I_{ACD},I_{BCD},I_{A,BC}$ $l_3: I_{B,AC},I_{B,CD},I_{A,CD},I_{A,BD}$ $l_4: I_{C,BD},I_{C,BA},I_{D,BA},I_{D,AC}$ we show only the first statement. For this we show that $\angle{I_{C,AD}I_{ABd}I_{ACD}}=90$. This angle is $\angle{AI_{ABd}I_{ACD}}-\angle{AI_{ABd}I_{C,AD}}=180-\angle{ADC}-\angle{ADI_{C,AD}}=180-90=90$. Now we prove that these four lines are parallel. We have shown $l_1||l_2$. For others we prove $l_1||l_3$(the other is the same). Because $I_{A,CD},I_{B,CD},I_{ACD},I_{BCD}$ are circumscribed,It's enough to show $\angle{I_{B,CD},I_{ACD},I_{BCD}}=\angle{I_{A,CD},I_{ACD},I_{BCD}}$. This two are equal to $180-\angle{ACD}/2-\angle{BDC}/2$

Define $M_{AB}$ to be the midpoint of arc $AB$, $I_A, I_B, I_C, I_D$ to be the four incenters, and $I_{AB}$ to be the $B$-excenter in triangle $BCD$, and so on. First, note that $I_A, I_B, I_{AB}, I_{BA}$ forms a rectangle as $M_{CD}I_{AB} = M_{CD} I_{BA} = M_{CD}I_A = M_{CD}I_B$. So the result follows for all but $I_{AC}$ and cyclic permutations. This follows by noticing that $M_{AB}$ is the center of $(I_{CD}BI_ADI_{CB})$, so $I_{AC}I_{CD}I_AI_{CB}$ is a rectangle too.

It's a known China TST problem that the four incentres form a rectangle. Now let $I$ and $J$ be the incentres of $\Delta ABC, \Delta ABD$. Then note that the $C-$ and $D-$ excentres of said triangles form a line parallel to $IJ$, because of the arc midpoint being a bisection point between the incentre and excentre. Thus we can easily find $\mathcal{K}$ and $\mathcal{L}$.