I think you meant incircles of $QAC$ and $QBC$

We know that $AI$ is perpendicular bisector of $PT$ and $BI$ is perpendicular bisector of $DF$ where $I$ is incenter of $\triangle ABC$ $AQ=s-a$, $BQ=s-b$ where $s$ is semiperimeter of $\triangle ABC$ In $\triangle AQC$, $PQ = \frac{AQ+QC-AC}{2} = \frac{1}{2} (QC+s-a-b)$ In $\triangle BQC$, $DQ = \frac{BQ+QC-BC}{2} = \frac{1}{2} (QC+s-b-a)$ Hence, $DQ=PQ$ As $IQ \perp AC$, $IQ$ is perpendicular bisector of $DP$ Hence, $I$ is point of concurrence of perpendicular bisectors of $PT, PD$ and $DF$ Hence, $TI=PI=DI=FI$ Hence, quadrilateral $PDFT$ is cyclic with $I$ as its center.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -71.19185509952509, xmax = 16.663997887308277, ymin = -25.576233484057415, ymax = 29.319319101179303; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); /* draw figures */ draw((-55.085904387159786,-2.158207995059887)--(-17.974886648735875,-11.618904066404488), linewidth(0.8) + blue); draw((-17.974886648735875,-11.618904066404488)--(-22.467621704207588,15.841352622475867), linewidth(0.8) + blue); draw((-22.467621704207588,15.841352622475867)--(-55.085904387159786,-2.158207995059887), linewidth(0.8) + blue); draw(circle((-29.62810144443004,1.100577031115594), 9.446600537626408), linewidth(0.8) + red); draw((-31.96168415191031,-8.053255688834701)--(-22.467621704207588,15.841352622475867), linewidth(0.8) + blue); draw(circle((-36.199800158298196,0.25935151713595417), 7.008047110875394), linewidth(0.8) + qqwuqq); draw(circle((-24.654725547453083,-4.327861543262558), 5.414966671273899), linewidth(0.8) + qqwuqq); draw(circle((-29.628101073122473,1.1005773471608873), 11.27770399068824), linewidth(0.8) + red); draw((-29.62810144443004,1.100577031115594)--(-31.96168415191031,-8.053255688834701), linewidth(0.8) + blue); draw((-29.62810144443004,1.100577031115594)--(-39.58570002655798,6.395181240194393), linewidth(0.8) + blue); draw((-29.62810144443004,1.100577031115594)--(-37.93098931234679,-6.5315032465262615), linewidth(0.8) + blue); draw((-29.62810144443004,1.100577031115594)--(-25.992378445480863,-9.5750082703329), linewidth(0.8) + blue); draw((-29.62810144443004,1.100577031115594)--(-19.310809003563335,-3.4535503727124843), linewidth(0.8) + blue); draw((-55.085904387159786,-2.158207995059887)--(-29.62810144443004,1.100577031115594), linewidth(0.8) + blue); draw((-29.62810144443004,1.100577031115594)--(-17.974886648735875,-11.618904066404488), linewidth(0.8) + blue); draw((-39.58570002655798,6.395181240194393)--(-37.93098931234679,-6.5315032465262615), linewidth(0.8) + blue); draw((-19.310809003563335,-3.4535503727124843)--(-25.992378445480863,-9.5750082703329), linewidth(0.8) + blue); /* dots and labels */ dot((-55.085904387159786,-2.158207995059887),dotstyle); label("$A$", (-54.88397127713248,-1.5737847595501469), NE * labelscalefactor); dot((-17.974886648735875,-11.618904066404488),dotstyle); label("$B$", (-17.731855667667656,-11.048435571855647), NE * labelscalefactor); dot((-22.467621704207588,15.841352622475867),dotstyle); label("$C$", (-22.210781506212104,16.399340720762712), NE * labelscalefactor); dot((-31.96168415191031,-8.053255688834701),linewidth(4pt) + dotstyle); label("$Q$", (-31.742854444652846,-7.603108003744556), NE * labelscalefactor); dot((-37.93098931234679,-6.5315032465262615),linewidth(4pt) + dotstyle); label("$P$", (-37.71475556271211,-6.052710598094565), NE * labelscalefactor); dot((-39.58570002655798,6.395181240194393),linewidth(4pt) + dotstyle); label("$T$", (-39.379997220632475,6.867267782322027), NE * labelscalefactor); dot((-25.992378445480863,-9.5750082703329),linewidth(4pt) + dotstyle); label("$D$", (-25.770953326593585,-9.096083283259363), NE * labelscalefactor); dot((-19.310809003563335,-3.4535503727124843),linewidth(4pt) + dotstyle); label("$F$", (-19.052564568776916,-3.009337912929768), NE * labelscalefactor); dot((-29.62810144443004,1.100577031115594),linewidth(4pt) + dotstyle); label("$I$", (-29.38854727311025,1.58443217788502), NE * labelscalefactor); dot((-29.687014437362116,-2.328378465188342),linewidth(4pt) + dotstyle); label("$E$", (-29.44596939924544,-1.8608953902260712), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Posting because of the lemma $\color{black}\rule{25cm}{1pt}$ We start off with this easy to prove lemma. $\color{red}\rule{25cm}{0.5pt}$ Lemma: The incircles of $QAC$ and $QBC$ touch each other at the same point on line $CQ$. Proof: Let's denote with $a,b,c$ the sides $BC,CA,AB$, respectively, and let's use $x,y,z$ for $AQ$,$BQ$ and $CQ$, respectively. We have that $x=s-a$ and $y=s-b$, where $s$ is the half perimeter of $ABC$. Now let's see that $CT=\frac{x+b+z}{2}-x$ and that $CF=\frac{a+z+y}{2}-y$. Now subtract $CT$ and $CF$ from which we have that: $$CT-CF=\frac{b-a+y-x}{2}$$but notice that the numerator is equal to $0$ this implies that $CT=CF$ and from power of a point we conclude that they touch at the same point on $CQ$, which we denote with $E$. $\color{red}\rule{25cm}{0.5pt}$ Now let's return to our problem at hand. From our lemma we have that $PQ=QE=QD$, which implies that $IP=ID$, where $I$ is the incenter of $ABC$. We easily notice that $ATI \cong API$ ($AI=AI$,$AP=AT$,$\angle PAI=\angle TAI$), this implies that $IP=IT$. Similarly we have that $BDI \cong BFI$, which implies that $ID=IF$. Thus we have that $IT=IP=ID=IF$, which implies that $TPDF$ is a cyclic quad whose circle has a center at the incenter of $ABC$.