Let the line passing through $D$ and parallel to $AB$ intersects $BP$ and $AT$ at $K$ and $L$, respectively. We have $$\frac{|DK|}{|AB|}=\frac{|PD|}{|PA|}=\frac{|PD|}{|PA|}\cdot\frac{|PT|}{|DL|}\cdot\frac{|DL|}{|PT|}=\frac{|PD|}{|PA|}\cdot\frac{|AP|}{|AD|}\cdot\frac{|DL|}{|PT|}=$$$$\frac{|PD|}{|AD|}\cdot\frac{|DL|}{|PT|}=\frac{|PT|}{|AB|}\cdot\frac{|DL|}{|PT|}=\frac{|DL|}{|AB|}\Rightarrow |DK|=|DL|$$Also, $\angle DCK=\angle ABC=\angle DKC\Rightarrow |DK|=|DC|$. Therefore, $\angle LCK=90^{\circ}$. Hence, it suffices to prove that the angle bisector of $\angle ACT$ is $CL$. Let $TA\cap PB=M$. We have $$\frac{|MA|}{|MT|}=\frac{|AB|}{|TP|}=\frac{|AD|}{|DP|}=\frac{|LA|}{|LT|}\Rightarrow (M,L;A,T)=-1$$Also, $\angle LCM=90^{\circ}$. Hence, we conclude that the angle bisector of $\angle ACT$ is $CL$.

Let $AB \cap CD = Q$ and $TC \cap AB = E$. Note that by the given angle condition, $\triangle QBC$ is isosceles with $QB = QC$. By angle chasing, it suffices to prove $\angle QCA = \angle CEQ$, or that $QC^2 = QA \cdot QE$. By Menelaus and similarity of triangles $\triangle BCE \sim \triangle PCT$ and $\triangle ADB \sim \triangle PDT$, we have $$1 = \dfrac{QA}{QB} \cdot \dfrac{BC}{CP} \cdot \dfrac{PD}{DA} = \dfrac{QA}{QB} \cdot \dfrac{BE}{PT} \cdot \dfrac{PT}{AB} \implies QA \cdot BE = QB \cdot AB.$$ Using the fact that $BE = QE - QB$, this gives $QA \cdot QE = QB^2 = QC^2$, which implies the result. $\square$

Let $TA\cap BC=G, TA\cap CF=K$, foot of the altitude from $A$ to $BC$ be $H$, the perpendicular at $C$ to $BC$ meet $BA$ at $F$, foot of the altitude from $T$ to $BC$ be $L$, $BA\cap CD=E, TC\cap AB=R$, $GD$ meet $BF, TB$ at $M,N$ respectively. We want to show that $BC$ is the angle bisector of $\angle RCA \iff (R,A;B,F)=-1$ \[(R,A;B,F)=(TC,TA;TB,TF)=(C,G;B,TF\cap BC)=(FC,FG;FB,FT)=(K,G;A,T)=(C,G;H,L)\]\[\frac{GH}{GL}=\frac{AH}{TL}=\frac{AB}{TP}=\frac{GB}{GP}=\frac{GM}{GN}=\frac{DM}{DN}=\frac{CH}{CL}\]As desired.$\blacksquare$

$AB\cap CD=Q$ We will use method of moving points. Take $QBCD$ fixed. Animate $A$ over $QB$. Denote $R(XY,ZT)$ as reflection of $XY$ to $ZT$. \[f:A\rightarrow AC\rightarrow R(AC,BC)\rightarrow R(AC,BC)\cap BQ\]\[g: A\rightarrow AD\rightarrow AD\cap BC=T\rightarrow T(QB)_{\infty}\cap BD=P\rightarrow PC\cap QB\]$f,g$ has degree $2$. $i)A=B$ $f: B\rightarrow BC\rightarrow BC\rightarrow B$ $g: B\rightarrow BD\rightarrow B\rightarrow B\rightarrow B$ So they are same at $A=B$. $ii)A=Q$ Let the parallel line from $C$ to $QB$ intersect $BD$ at $S$. $f:Q\rightarrow QC\rightarrow Q'C\rightarrow QB_{\infty}$ $g: Q\rightarrow QD\rightarrow C\rightarrow C(QB)_{\infty}\cap BD=S\rightarrow SC\cap QB=QB_{\infty}$ So they are same at $A=Q$. Let the parallel from $D$ to $BC$ intersect $QB$ at $E$ and the parallel from $C$ to $BD$ intersect $QB$ at $F$. $iii)A=E$ $f: E\rightarrow EC\rightarrow FC\rightarrow F$ $g: E\rightarrow ED\rightarrow BC_{\infty}\rightarrow (BC)_{\infty}(QB)_{\infty}\cap BD=BD_{\infty}\rightarrow F$ Thus $f,g$ are same at $3$ points as desired.$\blacksquare$