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)
Alternatively, Let $\angle A = 2a$, $\angle B = 2b$ with (wlog) $a \ge b$.
Then $\angle ABI= b$ ($BI$ is the bisector of $\angle ABC$), so $\angle ANI = b$ (angles subtended at circumference by same arc are equal).
Also, $\angle IAC = a$ ($AI$ is the bisector of $\angle BAC$), so $\angle AIN = a - b$ (external angle of triangle is sum of interior opposite angles).
Similarly, $\angle MBI= b$ ($BI$ is the bisector of $\angle ABC$), so $\angle MAI = b$ (angles subtended at circumference by same arc are equal).
Also, $\angle IAB = a$ ($AI$ is the bisector of $\angle BAC$), so $\angle BAM = a - b$ (subtraction).
Hence chords $AN$, $BM$ subtend equal angles at the circumference of the same circle, so are equal.
Best Answer
Notations. Let in $\triangle ABC$, $I$ be the incenter, $I_A,I_B,I_C$ be respective excenters. Let $\{AI\cap BC,BI\cap AC,CI\cap AB\}\equiv\{J_A,J_B,J_C\}$. Let $J_AJ_B\cap \odot(ABC)=\{L_C, L'_C\}$, $J_AJ_C\cap \odot(ABC)=\{L_B, L'_B\}$ and $J_CJ_B\cap \odot(ABC)=\{L_A, L'_A\}$.
Proof. Thus, we need to show that $IL_CL'_CI_AI_B$ is a cyclic pentagon. Clearly, its enough to show that $IL_CL'_CI_A$ is cyclic quadrilateral as the other part follows by symmetry. From incenter-excenter lemma, we know that $IBI_AC$ is cyclic and thus, $$J_AI\cdot J_AI_A\overset{\mathcal P_{\odot(IBC)}(J_A)}{=} J_AB\cdot J_AC\overset{\mathcal P_{\odot(ABC)}(J_A)}{=} J_AL_C\cdot J_AL'_C$$and thus, by converse of power of point theorem, we get $IL_CL'_CI_A$ cyclic which completes the proof. $\square$