= Solution
Let $T_y=\inf\{s\geq0:B_s=y\}$. Reflect at this <stopping time>, using the <Brownian reflection principle> from (a). Path <continuity> gives $\{S_t\geq y\}=\{T_y\leq t\}$, and on this <event> the reflected endpoint is $\widehat B_t=2y-B_t$. Consequently
$$
\{S_t\geq y,\ B_t\leq x\}
=\{T_y\leq t,\ \widehat B_t\geq2y-x\}.
$$
Since $x\leq y$, an endpoint $\widehat B_t\geq2y-x\geq y$ forces the reflected path to have hit $y$ by time $t$. Reflection preserves the first hit of $y$, so the condition $T_y\leq t$ is redundant on the right. As $\widehat B$ has the same <probability distribution> as $B$, the <joint distribution of Brownian motion and its running maximum> satisfies
$$
\boxed{\mathbb P(S_t\geq y,\ B_t\leq x)=\mathbb P(B_t\geq2y-x).}
$$
This argument includes $x=y$; at positive $t$ the boundary endpoint has zero probability under the <normal distribution>. At $t=0$ both sides vanish because $y>0$.
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