We treat the whole thing as an absorbing Markov chain.

The state transition matrix is

\(\displaystyle T = \begin{pmatrix}

0 &\dfrac 1 2 &\dfrac 1 2 &0\\

\dfrac 5 6 &0 &0 &\dfrac 1 6\\

0 &0 &1 &0\\

0 &0 &0 &1

\end{pmatrix}\)

This breaks up into the $Q$ and $R$ matrices as

$Q=\begin{pmatrix}0 &\dfrac 1 2 \\\dfrac 5 6 &0\end{pmatrix}\\

R = \begin{pmatrix}\dfrac 1 2 &0 \\0 &\dfrac 1 6\end{pmatrix}\\

N = \left(I_2 - Q\right)^{-1} =

\begin{pmatrix}

\dfrac{12}{7} & \dfrac{6}{7} \\

\dfrac{10}{7} & \dfrac{12}{7} \\

\end{pmatrix}\\

B = NR =

\begin{pmatrix}

\dfrac{6}{7} & \dfrac{1}{7} \\

\dfrac{5}{7} & \dfrac{2}{7} \\

\end{pmatrix}$

From $B$ we can directly read off that $P[\text{MW|start with MP}] = \dfrac 6 7$ and $P[\text{MW|start with SP}] = \dfrac 5 7$