Hint: Lemma: If $M$ is the midpoint of the arc $BAC$, $|b-c|/a=MA/MB$

Sketch of my solution. Let $A'$ be the $A$ - antipode wrt $(ABC)$. Ptolemy's thm on $CBPD$ along with $PD=\frac{KC\cdot KA'}{KA}$ and then we're done after a straightforward trig bash on the equation $\overline{DB}-\overline{DC}=\overline{AB}-\overline{AC}$.

Sketch of my solution: Let $P'$ be the $P$ - antipode wrt $(ABC)$. Let $R$: intersection between $AP$ and $P'D$. By angles chasing: $RCDK$ is cyclic, so $RL$ is the bisector of angle $C$ of triangle $ACD $. Finally, $R$ is the intersection of three bisectors of $ACDB$, so $ABDC$ is circumscribed.

Choose point $E$ on ray $BD$ s.t. $BE=CD$ . Note that $ABDC$ is circumscribed iff $AB+CD=AC+BD$ by Pitot's thm ie iff $BD-CD=AB-AC$. This is what we'll prove. Let $A,B,C$ denote angles $\angle BAC,\angle ABC$ and $\angle BCA$ respectively. Let $R$ be circumradius of $\triangle ABC$ Note that $\triangle PEB\cong\triangle PDC$ by SAS test. Hence $PD=PE$ ie $E$ lies on circle with centre $P$ and radius $PK$. $DE=BD-BE=BD-CD$ Now $\angle PDE=\angle PDB=\angle PAB= \frac{A}{2}$. $PD=PE$. So $\angle PED=\frac{A}{2}$. So $\angle EPD=180^{\circ}-A$. Hence by sine rule $DE=2PKsin\left(\frac{\angle EPD}{2} \right)=2PKsin\left(90^{\circ}-\frac{A}{2}\right)=2PKcos\left(\frac{A}{2}\right)$ Note that $\frac{CK}{CA}=tan\left(\frac{CAK}{2}\right)=tan\left(\frac{A}{2}\right)$. Hence $CK=CAtan\left(\frac{A}{2}\right)=2Rsin(B)tan\left(\frac{A}{2}\right)$. Now applying sine rule in $\triangle CKP$, $PK=\frac{CKsin(\angle PCK)}{sin(\angle CPK)}=\frac{CKsin\left(\frac{C-B}{2}\right)}{sin(B)}=2Rtan\left(\frac{A}{2}\right)sin\left(\frac{C-B}{2}\right)$. So $DE=4Rsin\left(\frac{A}{2}\right)sin\left(\frac{C-B}{2}\right)$. But now $sin(C)-sin(B)=2sin\left(\frac{C-B}{2}\right)cos\left(\frac{B+C}{2}\right)=2sin\left(\frac{C-B}{2}\right)sin\left(\frac{A}{2}\right)$. Hence $DE=2R(sin(C)-sin(B))=AB-AC$ $\therefore BD-CD=DE=AB-AC$ as desired. $\textbf{Q.E.D}$

Well-known lemma. Let $K,L$ be points on the angle bisector of $\angle BAC$, so that $\angle LBA=\angle ACK=90^{\circ}$. Then, the midpoint of $KL$ is the midpoint of arc $BC$ of $(ABC)$. Proof. Let $P$ be the midpoint of arc $BC$ of $(ABC)$, we claim that $PL=PK$. Indeed, note that as $BL,CK$ concur on $(ABC)$ at $A$-antipode, $A'$, we get that $\triangle A'KL$ is isosceles and as $\angle A'PA=90^{\circ}$, we conclude $PK=PL$. $\square$ Wlog $AC>AB$ and $L$ be the point on $AP$ so that $\angle LBA=90^{\circ}$. We bash the equality of Pitot's theorem out to show desired incircle. To begin with, note that $$DP=\frac{KL}{2}=\frac{AK-AL}{2}=\frac{AC-AB}{2\cos{\angle \frac{A}{2}}}.$$We would like to show that \begin{align*} AB+CD&=AC+BD\Longleftrightarrow\\ \frac{AB-AC}{2R}&=\sin{\angle BAD}-\angle{DAC} \Longleftrightarrow\\ \frac{AB-AC}{2R}&=\sin{(\angle BAP-\angle DAP)}-\sin{(\angle{BAP}+\angle{DAP})} \\&=-2\cos{\angle\frac{A}{2}\sin{DAP}}, \end{align*}which is true. We are done.