For a fixed quadrilateral $ABCD$ as described, we first show that $m$, the minimum possible length of $CP + PQ + QC$, equals the length of $AC$. Reflect $B$, $C$, and $P$ across line $AD$ to points $E$, $F$, and $R$, respectively, and then reflect $D$ and $F$ across $AE$ to points $G$ and $H$, respectively. These two reflections combine to give a $60^{\circ}$ rotation around $A$, so triangle $ACH$ is equilateral. It also follows that $RH$ is a $60^{\circ}$ rotation of $PC$ around $A$, so, in particular, these segments have the same length. Because $QR = QP$ by reflection,
$$CP + PQ + QC = CQ + QR + RH$$
The latter is the length of a broken path $CQRH$ from $C$ to $H$, and by the "shortest path is a straight line" principle, this total length is at least as long as $CH = CA$. (More directly, this follows from the triangle inequality: $(CQ + QR) + RH \geq CR + RH \geq CH$.) Therefore, the lower bound $m \geq AC$ indeed holds. To see that this is actually an equality, note that choosing $Q$ as the intersection of segment $CH$ with ray $AD$, and choosing $P$ so that its reflection $R$ is the intersection of $CH$ with ray $AE$, aligns path $CQRH$ with segment $CH$, thus obtaining the desired minimum $m = AC$.

We may conclude that the largest possible value of $m$ is the largest possible length of $AC$, namely $2$: the length of a diameter of the circle.