my solution (may be it is complicated): Let K' on AB AK'=BK Let K'' on line AB,AK''=AK' Then we have $\angle{P'K''K}=\angle{PTB}$ So need to prove P'K''K similar to PTB sufficient to provePK''/KK''=PT/TB that is PK/AB=PT/TB by PTK simialr to BTA done.

$PTB\sim TKA$ so $PT/TK=PB/KA=AP'/AK$ so $P'AK\sim PTK$ done.

we need to prove that $AT$ is parallel to $PK$ and because $A$ is midpoint of $PP'$ then we need to prove that $AT$ goes through midpoint of $PK$ so now let $AT$ intersect $PK$ at $L$ then we have $ \angle LKT= \angle PBT= \angle TAK \Rightarrow LK^2=LT.LA$ in same way we can prove that $LP^2=LT.LA $ so now $LP=LK$ and we are done

lemma:suppose a tiangle ABC and point T on the BC. so we would have: BT/CT=(sin(BAT)*BA)/(sin(CAT)*CA) in the triangle AKP , continue the line AT to meet PK at the point S.since the PTKB is embedded in a circle we will easily find it out that $\angle TPK=\angle TAP $ and $\angle TKP = \angle KAT $ now use the lemma in two triangles $%Error. "triagles" is a bad command. AKP $ and $%Error. "triagles" is a bad command. TKP $ for the point S and we will prove that SP=PK. so the two lines AT and kp' are parallel and $\angle P'KA=\angle KAT =\angle TBP$ and we are done

First , $AT \cap \circ BKP$ =$ {X} $ where $X$ is distinct from $T$ and also let $AT \cap KP=Z$. By an angle chase $AKXP$ is a $||^{gm}$.Diagonals of a $||^{gm}$ bisect each other and $P{P}'$ is bisected at $A$. So , $ZA || K{P}'$ which (again after an angle chase) yields us \[ \angle PBT = \angle {P}'KA \]

Restated Iran Second Round 2013 P4 wrote: Let $P$ be a point out of circle $\omega$. Let $PA$ and $PB$ be the tangents to the circle drawn from $C$. Choose a point $C$ on $AB$ . Suppose that the circumcircle of triangle $PBC$ intersects $\omega$ again at $K$. Let $Q$ be the reflection of $P$ with respect to $A$. Prove that \[ \angle DBP = \angle QCA \] Angle chase. $$\measuredangle PAD=\measuredangle ABD=\measuredangle CBD=\measuredangle CPD$$and $$\measuredangle DAC=\measuredangle DAB=\measuredangle DBP=\measuredangle DCP.$$ By this, we conclude that $D$ is the $A$-Humpty point of $\triangle APC$. By Humpty properties, we have $AD$ bisecting $PC$, hence $AD\parallel CQ$. Thus, $$\measuredangle DBP=\measuredangle DAB=\measuredangle DAC=\measuredangle QCA$$

step 1: TAK and TBP are similar. well easy to see ∠TAK = ∠PBT and ∠TKA = ∠TPB. step 2: AP'K and TBA are similar. from step1 we have AT/AK = BT/BP = BT/AP = BT/AP' and ∠P'AK = 180 - ∠BAP = 180 - ∠ABP = ∠KTP = ∠ATB. now from step2 we have ∠P'KA = ∠BAT = ∠PBK. we're Done.