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For $t>0$, write $\Phi$ for the standard normal distribution function. Since the Brownian running maximum is nonnegative, $a\le 0$ gives simply $\mathbb{P}(B_t\le b)=\Phi (b/\sqrt{t})$.

For $a>0$ and $b<a$, reflect the Brownian motion after its first hit of $a$. This is a stopping time, and the Strong Markov property together with symmetry of Brownian increments shows that reflection preserves its law. On paths that hit $a$, it sends the endpoint $B_t$ to $2a-B_t$. Thus the event with endpoint at most $b$ maps to endpoints at least $2a-b$, all of which necessarily hit $a$. This provesThe same reflection gives $\mathbb{P}(M_t\ge a)=2(1-\Phi (a/\sqrt{t}))$. If $b\ge a$, every path with $B_t>b$ has hit $a$, so subtracting that endpoint tail gives the complete answer:The expressions agree at $b=a$. This is the joint distribution of Brownian motion and its running maximum. In particular $M_t$ has the distribution of $|B_t|$ and has no atoms for $t>0$. If $t=0$, the pair is $(0,0)$ deterministically, so the requested probability is $1_{\{a\le 0,b\ge 0\}}$.

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

On the interior of the support $b<m$, differentiating the joint distribution of Brownian motion and its running maximum gives the joint probability densityMake the change-of-variables formula for a probability density $y=m$ and $z=m-b$. Its inverse is $m=y$, $b=y-z$, and its Jacobian determinant has absolute value one. Since $2m-b=y+z$, for $y,z\ge 0$ this yields