Yes, Ceva's a good choice here. Let's recall another version of Ceva's theorem
In the above picture, the three lines $XI, YH, ZG$ are concurrent if and only if
$$\frac{YI}{IZ}\cdot \frac{ZH}{HX} \cdot\frac{XG}{GY} = 1.$$
Now, the ratios of the lengths are equal to the ratios of the areas of certain triangles, more specifically:
$$\frac{YI}{IZ} = \frac{\triangle XYI}{\triangle XZI},\dots$$
here, we simply denote the triangle by its area. But
$$\triangle XYI = \frac12 XY\cdot XI \cdot \sin (\angle YXI), $$
and similar for $\triangle XZI$. So
$$\frac{YI}{IZ} = \frac{XY}{XZ}\cdot \frac{\sin\angle YXI}{\sin \angle ZXI}.$$
Taking cyclic product, we have the sine version of Ceva's theorem: That the three lines $XI, YH, ZG$ are concurrent if and only if
the product $\frac{\sin\angle YXI}{\sin \angle ZXI} \cdot \frac{\sin\angle ZYH}{\sin \angle XYH} \cdot \frac{\sin\angle XZG}{\sin \angle YZG}$ of cyclic ratios of sines is equal to $1$.
Now that's what we will use in this problem:
So we want to show that
$$\mathcal{F} = \frac{\sin\angle BAP}{\sin\angle PAC}\cdot \frac{\sin\angle ACR}{\sin\angle RCB} \cdot \frac{\sin\angle{CBQ}}{\sin\angle QBA} = 1.\tag{1}$$
Now
$$\frac{\sin\angle BAP}{\sin\angle PAC} \cdot \frac{AF}{AE} =\frac{\triangle AFP}{\triangle AEP} = \frac{FP}{PE},$$
and so on.
Taking the cyclic product of the above, we obtain
$$\mathcal{F} \cdot \frac{AF}{AE}\cdot\frac{BD}{BF}\cdot \frac{CE}{CD} = \frac{FP}{PE}\cdot\frac{ER}{RD}\cdot\frac{DQ}{QF}.$$
And (1) follows from classic Ceva's theorem with the given concurrences.
Suggested approach: Prove that $TF \times TX = TB \times TA$, where lines $DE$ and $AB$ intersect at point $T$. (In particular, independent of the $P,Q$, as suggested by the problem.)
Corollary: It follows that $TF \times TX = TB \times TA = TP \times TQ$ and thus $F,X,P,Q$ are concyclic as desired.
Proof of approach: One way to prove the equation is by side length chasing. Apply Menelaus on triangle $ABC$ to transversal $TDE$ to obtain $TA/TB$ and hence $TA, TB$. Then we can find $TF, TX$ and multiply it out.
Details of side length chasing
$\frac{AT}{TB} \times \frac{BD}{DC} \times \frac{CE}{EA} = 1 $
$ \frac{AT} {TB} = \frac{ EA}{BD} = \frac{c+b-a}{c+a-b}$
$AT - TB = c \Rightarrow AT = \frac{ c (c+b-a) } { 2(b-a) } , TB = \frac{ c(c+a - b ) } { 2(b-a)}$
$TF = TB + BF = \frac{ c(c+a - b ) } { 2(b-a)} + \frac{c+a - b}{2} = \frac{ ((c-b+a)(c+a-b) } { 2(b-a)} $
$TX = TB + BX = \frac{ c(c+a - b ) } { 2(b-a)} + \frac{c}{2} = \frac{ c(c)}{ 2(b-a)}$
Now, we multiply these terms to show that $TA \times TB = \frac{ c^2 (c+b-a)(c+a-b) } { 4(b-a)^2} = TX \times TF$
Additional observations, which I couldn't use directly
$A, F, T, B$ are harmonic conjugates. This can be shown either from
1) $ \frac{ TA}{TB} = \frac{EA}{BD} = \frac{AF}{FB}$, or also from
2) Lines $AD, BE, CF$ are concurrent (at the Gergonne point)
Best Answer
Note $BZ\cdot BZ’=BX\cdot BX’$, etc. by power of a point so we’re done by Ceva twice.