Similar to (Actually the same as) MEMO 2021:

Let the tangent at $A'$ intersect $BC$ at $D$, $DA''$ be tangent and $E'=DO\cap AB$, $F'=DO\cap AC$, $P'=DO\cap A'A''$, $BC\cap A'A''=X$.

$\angle AA''A'=\angle E'P'A'=90^{\circ}\Rightarrow AA''\mid\mid E'F'\Rightarrow (E' ,F' ;O , P_{\infty})\overset{A}{=}( B, C; A', A'')\overset{A''}{=}(B , C; X, D)=-1\Rightarrow OE'=OF'\Rightarrow E', F'=E, F\Rightarrow P'=P$. That means that $BC, A'D, EF$ concurr. Remark:Click to reveal hidden text

Now, $PA'$ bissects $\angle BPC$ because $\angle A'PD=90^{\circ}$ and $(B , C; X, D)=-1$. $_{\blacksquare}$

Attachments:

[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 = -13.603676519583704, xmax = 22.252541146712478, ymin = -8.639836094706249, ymax = 7.140634701137643; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); draw((-2.2064676625978508,-2.8228874524998453)--(-2.2056878230549017,-2.327689342727229)--(-2.7008859328275174,-2.3269095031842806)--(-2.7016657723704665,-2.8221076129568963)--cycle, linewidth(2) + qqwuqq); draw((2.798586377032626,-2.830769427365405)--(2.799366216575575,-2.3355713175927892)--(2.304168106802959,-2.33479147804984)--(2.30338826726001,-2.829989587822456)--cycle, linewidth(2) + qqwuqq); draw((7.803640416663105,-2.838651402230965)--(7.804420256206053,-2.3434532924583493)--(7.309222146433438,-2.3426734529154003)--(7.3084423068904885,-2.837871562688016)--cycle, linewidth(2) + qqwuqq); /* draw figures */ draw((-0.09720010190841756,5.455625726278238)--(-4.04,-2.82), linewidth(2)); draw((-0.09720010190841756,5.455625726278238)--(8.66,-2.84), linewidth(2)); draw(circle((2.3132443270799086,-0.769852304257943), 6.675838453252857), linewidth(2)); draw((-0.09720010190841756,5.455625726278238)--(4.704257935145613,-7.002818931793674), linewidth(2)); draw((4.704257935145613,-7.002818931793674)--(5.5941230192678,-1.2940482006684315), linewidth(2)); draw((-4.04,-2.82)--(15.527826191121685,-2.850815474316727), linewidth(2)); draw((4.704257935145613,-7.002818931793674)--(15.527826191121685,-2.850815474316727), linewidth(2)); draw((-2.6972240584960305,-0.001619302689852553)--(7.310452208321963,-1.5615841537015296), linewidth(2)); draw((-2.6972240584960305,-0.001619302689852553)--(-2.7016657723704665,-2.8221076129568963), linewidth(2)); draw((2.306614074912966,-0.7816017281956911)--(2.30338826726001,-2.829989587822456), linewidth(2)); draw((7.310452208321963,-1.5615841537015296)--(7.3084423068904885,-2.837871562688016), linewidth(2)); /* dots and labels */ dot((-0.09720010190841756,5.455625726278238),dotstyle); label("$A$", (0.0058123472461609395,5.6933134151282925), NE * labelscalefactor); dot((-4.04,-2.82),dotstyle); label("$B$", (-3.939305351714812,-2.5937681418607337), NE * labelscalefactor); dot((8.66,-2.84),dotstyle); label("$C$", (8.759771738431752,-2.617112033570562), NE * labelscalefactor); dot((-2.6972240584960305,-0.001619302689852553),dotstyle); label("$E$", (-2.608703524254602,0.2308427550284836), NE * labelscalefactor); dot((7.310452208321963,-1.5615841537015296),dotstyle); label("$F$", (7.405826019261714,-1.3331979895300088), NE * labelscalefactor); dot((2.306614074912966,-0.7816017281956911),linewidth(4pt) + dotstyle); label("$O$", (2.41023319335847,-0.5861934548155049), NE * labelscalefactor); dot((4.704257935145613,-7.002818931793674),linewidth(4pt) + dotstyle); label("$A'$", (4.791310147760951,-6.819012541339646), NE * labelscalefactor); dot((5.5941230192678,-1.2940482006684315),linewidth(4pt) + dotstyle); label("$P$", (5.678378032734424,-1.0997590724317263), NE * labelscalefactor); dot((15.527826191121685,-2.850815474316727),linewidth(4pt) + dotstyle); label("$T$", (15.622875901121256,-2.6637998169902186), NE * labelscalefactor); dot((-2.7016657723704665,-2.8221076129568963),linewidth(4pt) + dotstyle); label("$R$", (-2.608703524254602,-2.6404559252803903), NE * labelscalefactor); dot((2.30338826726001,-2.829989587822456),linewidth(4pt) + dotstyle); label("$M$", (2.3868893016486417,-2.6404559252803903), NE * labelscalefactor); dot((7.3084423068904885,-2.837871562688016),linewidth(4pt) + dotstyle); label("$Q$", (7.405826019261714,-2.6404559252803903), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] I will only test the concurrence. Since $T$ is the intersection of $BC$ and the tangent through $A'$ to $\Gamma$, we want to prove that $E,F,T$ are collinear, that is, by Menelaus' Theorem, we want $\dfrac{AE}{BE}\cdot\dfrac{BT}{CT}\cdot\dfrac{CF}{AF}=1$ $\Longrightarrow$ $\boxed{\dfrac{AE}{AF}\cdot \dfrac{BT}{CT}\cdot\dfrac{CF}{BE}}=1$ Calculating $\dfrac{AE}{AF}$: Since $\angle{BAC}=\alpha, \angle{ABC}=\beta, \angle{BCA}=\theta$, it is easy to see that $\angle{BAO}=90-\theta, \angle{CAO }=90-\beta$, therefore by the Ratio Lemma in $\triangle{AEF}$: $\dfrac{AE}{AF}\cdot\dfrac{cos\theta}{cos\beta}=1$ $\Longrightarrow$ $\boxed{\dfrac{AE}{AF}=\dfrac{cos\beta} {cos\theta}}$ Calculating $\dfrac{BT}{CT}$: Marking angles, we find $\angle{A'BC}=90-\beta, \angle{BCA'}=90-\theta, \angle{BA'C}=\beta+\theta, \angle{CA'T} =90-\beta$ (tangency), $\angle{A'CT}=90+\theta$. By Sinine Law in $\triangle{BA'T}$ and $\triangle{CA'T}$, respectively: $\dfrac{BT}{cos\theta}=\dfrac{A'T}{cos\beta}$ $\Longrightarrow$ $A'T=\dfrac{BT\cdot{cos\beta}}{cos\theta }$ $\dfrac{CT}{cos\beta}=\dfrac{A'T}{cos\theta}$ $\Longrightarrow$ $A'T=\dfrac{CT\cdot{cos\theta}}{cos\beta }$ $\Longrightarrow$ $\boxed{\dfrac{BT}{CT}=(\dfrac{cos\theta}{cos\beta})^2}$ Calculating $\dfrac{CF}{BE}$: Let $R,M,Q$ be the intersections of the perpendiculars to $BC$ by $E,O,F$, respectively, see that $REFQ$ is a trapezoid, and as $O$ is the midpoint of $EF$, ${ RM=MQ} (1.0)$. Since $O$ is the projection of the circumcenter onto $BC$, ${BM=MC} (1.1)$. For $(1.0)$ and $(1.1)$, $\boxed{BR=QC}.$ In the triangles $\triangle{BRE}$ and $\triangle{QFC}$, respectively: $cos\beta=\dfrac{BR}{BE}$ $cos\theta=\dfrac{QC}{CF}$ $\Longrightarrow$ $\boxed{\dfrac{cos\beta}{cos\theta}=\dfrac{BR\cdot{CF}}{QC\cdot{ BE}}={\dfrac{CF}{BE}}}$ Finally, $\dfrac{AE}{AF}\cdot\dfrac{BT}{CT}\cdot\dfrac{CF}{BE}=\dfrac{cos\beta}{cos\theta}\cdot(\dfrac {cos\theta}{cos\beta})^2\cdot\dfrac{cos\beta}{cos\theta}=1$ $Q.E.D!$