Note that $BP$ is parallel to $CQ$ so it suffices to prove that $\frac {BP} {CQ} =\frac{BD} {CD} $. but from the law of sin in $CQE$ we have $CQ=2CD. sin(\frac{A} {2})$ similarly $BP=2BDsin(\frac{A} {2})$.

Clearly $BP\parallel CQ$ so $\angle DBP=\angle DCQ$ now since $\Delta AFE\sim \Delta FBP\sim \Delta EQC$ so $\frac{BD}{BP}=\frac{BF}{BP}=\frac{QE}{QC}=\frac{DC}{CQ}$ so $\Delta DBP\sim \Delta DCQ$ so $\angle DPB=\angle DQC$ so we conclude that $D,P,Q$ are collinear $\blacksquare$

We will prove BPD and CQD are similar. ∠BFP = ∠BAC = ∠QEC so ∠PBF = ∠CQE so BP || CQ and ∠PBD = ∠QCD. we have to prove BP/BD = CQ/CD. BP/BD = BP/BF = CQ/CE = CQ/CD. we're Done.

Almost panicked before I noticed this was in fact straightforward. We start off by making the following observations. Claim : $\triangle FBP \sim \triangle EQC$ and lines $\overline{BP}$ and $\overline{CQ}$ are parallel. Proof : This is easy to see since both $\triangle FBP$ and $\triangle EQC$ are isosceles and their vertex angles are equal, \[\measuredangle PFB = \measuredangle CAB = \measuredangle CEQ\]Thus, they are indeed similar from which it follows that lines $\overline{BP}$ and $\overline{CQ}$ are parallel (since the other two pairs of sides $BF$ and $EQ$ and $FP$ and $EC$ are known to be parallel). Now, we note that \[\frac{BP}{CQ}=\frac{BF}{CE}=\frac{BD}{CD}\]which combined with the fact that \[\measuredangle PBD = \measuredangle PBC = \measuredangle QCB = \measuredangle QCD\]due to the above parallel lines, implies that $\triangle BPD \sim \triangle CQD$. Thus, $\measuredangle BDP = \measuredangle CDQ$ which implies that $P-D-Q$ as desired.

Sol:- Reflect $E$ across $CQ$ to get $E'$ . Observe that $D$ is homothetic center of $BFP$ and $CE'Q$,so the result follows.

Attachments:

We have $\angle BFP=\angle FAE=\angle QEC$, $BF=FP$, $FA=AE$ and $QE=EC$, so $BFP\simeq FAE\simeq QEC$ and $BP\parallel FE\parallel QC$. Hence$$\frac{BP}{QC}=\frac{BF}{CE}=\frac{BD}{DC}.$$We conclude that $P,D,Q$ are collinear, as wanted.

Using the fact that $FP$//$AC$ and $EQ$//$AB$, we get that, $\angle CEQ$=$\angle CAB$=$\angle PFB$. and as $CE$=$CQ$ and $PF$=$PB$, we can conclude that $\triangle CEQ$ $ \sim$ $PFB$. $\therefore$ $CQ$//$PB$. As $CE$,$CD$,$FB$,$DB$ are all tangents to the circle, $CE$=$CD$, $PF$=$FB$=$BD$. From similarity, $\frac{CQ}{CB}$=$\frac{CE}{PF}$=$\frac{CD}{BD}$, which implies that $\triangle CDQ$ $ \sim$ $\triangle PDB$. Hence, $\angle CDQ$=$\angle BDP$, from which we can conclude that $P$, $D$ and $Q$ are collinear.