Let $F_1$ and $F_2$ be the feet of perpendiculars from the points $J$ and $K$ to $AC$, respectively. Then
$$AF_1 = \frac{AB + AC - BC}{2} \quad \text{and} \quad AF_2 = \frac{AD + AC - CD}{2}.$$
Since $AB + CD = BC + AD$, we have $AF_1 = AF_2$. It follows that $F_1 = F_2$, that is, $JK \perp AC$.
Since $P$ is the circumcenter of $AJK$, we have $\angle PAK = 90^\circ - \angle AJK$ and since $J$ is the incenter of $ABC$ we have $\angle AJK = 90^\circ - \angle BAC/2$. From this we deduce that $\angle PAK = \angle BAC/2$. Hence $\angle PAD = \angle BAC/2 + \angle CAD/2 = \angle BAD/2$. It follows that the points $A, P, I$ are collinear. Similarly, the points $C, Q, I$ are collinear.
Moreover, since $PQ \parallel AC$ we have $\frac{AC}{PQ} = 1 + \frac{AP}{PI}$. On the other hand, we have $1 + \frac{PI}{AI} = \frac{r}{PN}$. Finally, because $PN = AP \sin \angle BAD/2 = PJ \sin \angle JAK = JK/2$ we have
$$\frac{AC}{PQ} = 1 + \frac{JK/2}{r - JK/2}.$$
This is equivalent to the statement of the problem.