Here is my partial solution: -Note that $\angle APS= \angle ACB = \angle BAT = \angle ATS$, so $ASTP$ is cyclic. -Next, $\angle CAP = \angle CBP = \angle SBT = \angle ASP$, so $(ASP)$ is tangent to $AC$. -We are left to prove $R$ lies on the same circle. Note that $\angle BTR =\angle SBT=\angle ASP = \angle ATP$, so $\angle ATB = \angle PTR$. But $\angle ATB = \angle SAT = \beta - \gamma$. We want $\angle PAR = \angle PTR$. We have $\angle PAR =\angle PCB = \angle ACP - \gamma$. So it's left $\angle ACP =\beta \iff AT||CP$. -For a proof for the last fact, see the first paragraph below.

This is a UK proposal by Sam Bealing (me). Let $N$ be the midpoint of $BS$ which, as $SABT$ is a parallelogram, is also the midpoint of $TA$. Using $ST \parallel AB \parallel MP$ we get: $$ \frac{NB}{BP}=\frac{1}{2} \cdot \frac{SB}{BP}=\frac{TB}{2 \cdot BM}=\frac{TB}{BC} $$which shows $TA \parallel CP$. Let $\Omega$ be the circle with diameter $OT$. As $\angle OMT=90^{\circ}=\angle TAO$ we have that $A,M$ lie on $\Omega$. We now show that $P$ lies on $\Omega$. As $TA \parallel CP$ and $TA$ is tangent to $\odot ABC$ we have $AP=AC$ so: $$ \angle TAP=\angle ACP=\angle CPA=\angle CBA=\angle TMP $$where in the last step we used $MP \parallel AB$. This shows $P$ lies on $\Omega$. Furthermore, this shows that $\angle OPT=90^{\circ}$ so $TP$ is also tangent to $\odot ABC$. Now we show that $R,S$ lie on $\Omega$ which shows that $\Omega$ is the circumcircle of $\triangle STR$. For $S$, using $ST \parallel AB$ and that $TA$ tangent to $\odot ABC$: $$ \angle TSP=\angle ABS=\angle ACP=\angle TAP $$and for $R$ taking homothety factor $2$ at $A$ takes $BN \rightarrow RT$ so $BN \parallel RT$ and hence: $$ \angle ART=\angle ABS=\angle TSP=\angle TAP=\angle APT $$where the last step follows from $TA=TP$ as they are both tangents. Finally, we observe that as $TA$ tangent to $\odot ABC$: $$ \angle TAC=180^{\circ}-\angle CBA=\angle ABT=\angle TSA $$which by alternate segment theorem means line $AC$ is tangent to $\Omega$ as required.

Let $K$ be midpoint of $SB$. Claim $: AST$ is tangent to $AC$. Proof $:$ Note that $\angle ATS = \angle TAB = \angle ACB = \angle 180 - \angle SAC$. we will prove $ASTR$ is cyclic. Claim $: ASTP$ is cyclic. Proof $:$ Note that $\angle ATS = \angle TAB = \angle ACB = \angle APB = \angle APS$. Claim $: AT || PC$. Proof $:$ Note that $\frac{KB}{BP} = \frac{SB}{2BP} = \frac{TB}{2BM} = \frac{TB}{BC}$. we will prove $ATRP$ is cyclic. Note that $\angle RTP = \angle TPS = \angle TAS = \angle ATB = \angle BCP = \angle BAP = \angle RAP$ so $ARSTPR$ is cyclic and $AST$ is tangent to $AC$ so $STR$ is tangent to $AC$.

Solved with NoctNight. Basically a bunch of angle-chasing. Let $D$ be the shared midpoint of $BS, AT$. Then since $ST \parallel AB \parallel MP$, we have $$\frac{DB}{BP} = \frac{SB}{2BP} = \frac{TB}{2BM} = \frac{TB}{BC}$$and so $DT \parallel PC$. Since $DT$ is the tangent to $(ABC)$ at $A$, $\angle ACP = 180^\circ - \angle TAC = \angle CPA$ so $\triangle APC$ is isosceles, and from $\angle PBC = \angle PBA - \angle CBA = 180^\circ - 2\angle CBA$, $\angle BMP = \angle BTS = \angle SAB = \angle CBA$ we have $\angle BPM = \angle CBA = \angle BMP$ so $BP = BM \implies BS = BT$, which implies $S$ and $T$ are reflections about the line through $B$ perpendicular to $AB$ as $ST \parallel AB$, so $STAR$ is an isosceles trapezium and hence cyclic. Let $O$ be the circumcenter of $\triangle ABC$. Notice $\angle TAO = \angle OMT = 90^\circ$ and so $A, M \in (TO)$. Then $\angle TAP = \angle ACP = \angle CBA = \angle BMP$ implies $P \in (TO) \implies \angle OPT = 90^\circ \implies TP$ is tangent to $(ABC)$. Finally we have $\angle TSP = \angle ABS = 180^\circ - \angle PBA = \angle ACP = \angle TAP$ so $S \in (TO)$; then $STAR$ being cyclic implies $R \in (TO)$. Now since $TA = TP$ by tangents, $AP = AC$ and $\angle TAP = \angle ACP$ we have $\triangle PAT \sim \triangle PCA$ so $\angle PAC = \angle PTA$ and $AC$ is tangent to $(STR) = (OT)$ as required. $\square$

Nice solution with squarc_rs3v2m, I must say this problem resembles 2017 IMOSL G3 (especially with the construction of tangent at $A$ to the circumcircle and using $\angle TAO=\angle TMO=90^{\circ}$), as well as 2017 IMOSL G5 with the three lines $ST, AR, MP$ sharing the same perpendicular bisector.

Neat problem! The key fact lies in proving $AP$ is the $A$ Symmedian in $\triangle ABC$. Let $N$ be the midpoint of $\overline{AT}$, noticing $\triangle BMP \sim \triangle BTS$ we have $\dfrac{BP}{BC} = \dfrac{BP}{2BM} = \dfrac{BS}{2BT} = \dfrac{BN}{BT}$ meaning $\triangle BNT \sim \triangle BPC$ and $$\measuredangle ATM = \measuredangle PCB = \measuredangle PAB = \measuredangle APM \implies P \in (ATM)$$letting $O$ be the circumcenter of $\triangle ABC$ we have $O \in (ATM) \implies \overline{OP} \perp \overline{TP}$ meaning $P$ is the intersection of $A$ Symmedian with $(ABC)$. The rest of the problem is just angle chasing, $\measuredangle TSP = \measuredangle TSB = \measuredangle ABS = \measuredangle ACP = \measuredangle TAP \implies S \in (ATPMO)$ and $\measuredangle TRA = \measuredangle TRB = \measuredangle BST = \measuredangle PAT$ (from above) $=\measuredangle TPA \implies R \in (ASTPMO)$ and $\measuredangle MTA = \measuredangle BCP = \measuredangle BAP = \measuredangle MAC$ proves the problem.

is there any solution with simple lemma ?

$TA\cap BS=Q, SA\cap (ABC)=K$ Claim: $TA\parallel PC$. Proof: $2.\frac{BQ}{BP}=\frac{BS}{BP}=\frac{BT}{BM}=\frac{BT}{BC}.2\implies \frac{BQ}{BP}=\frac{BT}{BC}\iff TA\parallel PC$ Claim: $A,S,T,P,R$ are cyclic. Proof: $\angle SPA=\angle BPA=\angle BCA=\angle BAT=\angle STA\implies S,P,T,A$ are cyclic. We have $\angle PAB=\angle PCB=\angle PCT=\angle ATC=\angle B-\angle C=\angle TAS=\angle TPS\implies TP$ is tangent to $(ABC)$. $\angle TRA=\angle BST=\angle SBA=\angle B=\angle TPA\implies T,R,P,A$ are cyclic. $\implies A,S,T,P,R$ are cyclic. So $TP^2=TA^2=TB.TC$. Let's invert around $T$ with radius $TA$. $\angle BCS^*=\angle BST=\angle SBA=\angle B=\angle BCK\implies B,K,S^*$ are collinear. $T,S,A,P,R$ are cyclic hence $S^*,A,P,R^*$ are collinear. $(TSR)\rightarrow R^*S^*=S^*APR^*$ $AC\rightarrow (TAB)$ $\angle PAB=\angle B-\angle C=\angle ATB$ therefore $(ATB)$ and $S^*APR^*$ are tangent to each other which gives the desired conclusion.$\blacksquare$