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 that $TA \parallel CP$.

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 $\Gamma$ we have that $AP = AC$, so
$$\angle TAP = \angle ACP = \angle CPA = \angle CBA = \angle TMP$$
where in the last step we used the fact that $MP \parallel AB$. This shows that $P$ lies on $\Omega$. Furthermore, this shows that $\angle OPT = 90^\circ$ and so $TP$ is also tangent to $\Gamma$.

Now we show that $R, S$ lie on $\Omega$ which would show that $\Omega$ is the circumcircle of triangle $STR$. For $S$, using $ST \parallel AB$ and that $TA$ tangent to $\Gamma$ we have
$$\angle TSP = \angle ABS = \angle ACP = \angle TAP.$$
For $R$, the homothety with factor 2 centred at $A$ takes $BN$ to $RT$. So $BN \parallel RT$ and hence
$$\angle ART = \angle ABS = \angle TAP = \angle APT,$$
where the last step follows from $TA = TP$ as they are both tangents to $\Gamma$.

Finally, we observe that as $TA$ tangent to $\Gamma$ then
$$\angle TAC = 180^{\circ} - \angle CBA = \angle ABT = \angle TSA$$
which, by the alternate segment theorem, means that line $AC$ is tangent to $\Omega$ as required.

We have
$$\angle APS = \angle ACB = \angle TAB = \angle ATS,$$
so $S, A, P, T$ are concyclic on a circle $\Omega$. We also have
$$\angle PAC = \angle PBC = \angle SBT = \angle PSA$$
so $AC$ is tangent to $\Omega$. It remains to prove that $R$ belongs on $\Omega$.

As in Solution 1 we have that $TA \parallel CP$. Then
$$\angle CPM = \angle ATS = \angle APS.$$
Since also $\angle BAP = \angle BCP$, then the triangles $APB$ and $CPM$ are similar. But then the triangles $BPC$ and $RAP$ are also similar as $\angle RAP = \angle BCP$ and
$$\frac{RA}{AP} = \frac{2BA}{AP} = \frac{2MC}{CP} = \frac{BC}{CP}.$$
It now follows that
$$\angle ARP = \angle PBC = \angle ASP$$
and therefore $R$ belongs to $\Omega$ as required.