We first prove that $DPTR$ is cyclic. Let $S$ be on $EF$ such that $PS \parallel BC$; then it suffices to show that $\angle{DRS} = 90^{\circ}$, or that $A, R, S$ are collinear. Note that $\triangle{PEF} \sim \triangle{A'BC} \implies$ if $J = BC \cap A'A'$ then $PEFS \sim A'BCJ \implies \angle{A'AS} = \angle{DAJ}$, so it suffices to show that $\angle{A'AR} = \angle{DAJ}$, which is equivalent to $R, J$ being inverses under force-overlay inversion at $A$. However, this is true because the inverse of $A'D$ is $(ADA')$, which passes through $J$ since $\angle{ADJ} = \angle{AA'J} = 90^{\circ}$. Thus, $A, S, R$ are collinear. Now we will show that $DQ$ is tangent to $(DPTR)$. Force-overlay inverting everything, $(DPTR)$ goes to $(A'P^*J)$ and $DQ$ goes to $(AA'Q^*)$, if $P^*, Q^*$ are the inverses of $P$ and $Q$, so it suffices to show that these two circles are tangent. If $X$ lies on the tangent to $(ABC)$ at $A'$ such that $Q^*X \parallel BC$, then by homothety it suffices to show that $\triangle{A'P^*J} \sim \triangle{A'AX}$. By angle chasing, $BPCQ^*$ is cyclic, and also $PA' \parallel DP^*$, so if $K = AD \cap (ABC)$ then this similarity is equivalent to $\frac{DK}{KQ^*} = \frac{DP}{PA}$, which is true by Power of a Point at $D$ onto $(ABC), (BPCQ^*)$. $\square$

Let $AR\cap EF = G$, since $\angle ARD= \angle GTD = 90^{\circ}$ it suffices to show $D,P,G,R$ concyclic. By angle chasing we get $B,R,G,F$ and $D,B,F,P$ concyclic and so $D,P,G,R$ are indeed concyclic by radical axis. Now we only need to prove $DQ$ is tangent to $\odot (DPGR)$. Let $\odot (DEF)$ meet $GD$ at another point $K$. Note that $\angle GPF= \angle FEP$, hence $GK \cdot GD = GF \cdot GE = GP^2$, which means $\angle PKD = \angle GPD = 90^{\circ}$, or that $P$ has a pedal circle with respect to the quadrilateral formed by lines $AB,BC,CA,GD$. But since $\angle ABP = \angle QBC$, $Q$ is exactly the isogonal conjugate of $P$ wrt. the above quadrilateral, so $\angle GDQ = \angle PDC = 90 ^ {\circ}$, and the desired tangency follows. $\blacksquare$

Nice problem, really liked it. Trivially $P,Q$ are isogonal conjugates, now let $AR \cap EF=G$, let $J,K$ be the inverses of $Q$ in $\sqrt{AP \cdot AD}$ and force overlay invert respectively, let $AA' \cap EF=H$ (note that by antiparalels we have $AA' \perp EF$). Claim 1: $DRTGP$ is cyclic. Proof: As $\angle GRD=90=\angle GTD$ we have $GTRD$ cyclic, but by $\sqrt{AP \cdot AD}$ invert we have $PGRD$ cyclic proving the claim. Claim 2: $PJA'KR$ is cyclic. Proof: First note that $\angle AKC=\angle ABQ=\angle PBC$ therefore $PBKC$ is cyclic (remember that from force-overlay we have $\triangle AKC \sim \triangle ABQ$), now by PoP: $$RD \cdot DA'=BD \cdot DC=PD \cdot DK \implies PRKA' \; \text{cyclic!}$$Now as $D \to A'$ in force overlay, we have $DQ \parallel KA'$ but also $PJQD$ is cyclic because of $\sqrt{AP \cdot AD}$ invert therefore by Reim's theorem we get $PJA'K$ cyclic, joining both cyclics implies the claim. The finish: Note that $H \to A'$ in $\sqrt{AP \cdot AD}$ invert so inverting back we get $GHQD$ cyclic therefore $\angle GDQ=90$ but $GD$ is the diameter of $(DPTR)$ therefore $DQ$ is tangent to $(DPTR)$ as desired, thus we are done .

Claim: $D,P,T,R$ are concyclic. Proof: Take $\sqrt{bc}$ inversion and reflect over the angle bisector of $\measuredangle A$. Inverted Claim: $ABC$ is a triangle with altitude $AD$ and circumcenter $O$. $A'$ is the antipode of $A$ on $(ABC)$.$(ADA')$ meets $BC$ at $R$. $P$ is an arbitrary point on $AO$ and the perpendicular line to $AO$ at $P$ intersects $AB,AC$ at $E,F$ respectively. If $(PA'R)$ meets $(AEF)$ at $K$, then prove that $(AKA')$ is tangent to $AD$. Since $\measuredangle OAR=\measuredangle AA'R=\measuredangle ADR,$ we get that $RA'$ is tangent to $(ABC)$. Under the inversion centered at $A$ with radius $\sqrt{AP.AA'},$ $(AEFK)$ swaps with $BC$ thus, $AK\cap BC=L$ is the image of $K$. We have $AK.AL=AP.AA'$ so $L\in (PKA'R)$. The diameter of this circle is $PR$ thus, $\measuredangle ALP=\measuredangle PLR=90$. \[\measuredangle KAD=\measuredangle LAD=\measuredangle ALP=\measuredangle KLP=\measuredangle KA'P=\measuredangle KA'A\]Hence $(AKA')$ is tangent to $AD$.$\square$ Lemma: $A',D$ are the anitpode of $A$ and the altitude from $A$ to $BC$ respectively. $P$ is on $AD$ and $Q$ is the isogonal conjugate of $P$. If $H$ is the altitude from $Q$ to $BC$, then show that $A',H,P$ are collinear. Proof: Note that $Q\in AA'$. \[\frac{QH}{AP}=\frac{\sin \measuredangle QBC.BQ}{\sin \measuredangle ABP.\frac{AB}{\sin 90-\measuredangle B+\measuredangle QBC}}=\frac{BQ}{AB}.\sin 90-\measuredangle B+\measuredangle QBC \]\[\frac{A'Q}{A'A}=\frac{\sin 90-\measuredangle B+\measuredangle QBC.\frac{BQ}{\sin \measuredangle C}}{\frac{AB}{\sin \measuredangle C}}=\frac{QB}{AB}.\sin 90-\measuredangle B+\measuredangle QBC\]Thus, $\frac{A'Q}{A'P}=\frac{HQ}{PA}$ and combining this with $AP\parallel QH$ implies the collinearity of $A',H,P$.$\square$ Claim: $(DPR)$ is tangent to $DQ$. Proof: Let $S$ be the antipode of $D$ on $(DPRT)$. $SP\perp AD$ and $R,S,A$ are collinear. Let $SP\cap A'D=M$. By Desargues on $\triangle HQD$ and $\triangle PAM,$ since they are centerally perspective, $PM\cap DH=BC_{\infty},AP\cap HQ=AD_{\infty}, DQ\cap AM$ are collinear. Hence $AM\parallel DQ$ as desired.$\blacksquare$