Conditioned Maximum of Brownian Motion

The joint distribution of $(B_t, M_t)$ is well-known. The probability density reads: $$ f_{(B_t, M_t)}(x,y) = \sqrt{\frac{2}{\pi}} \frac{2y-x}{t^{3/2}} \exp\left(-\frac{(2y-x)^2}{2t} \right) [ y>0, x \leqslant y ] $$ where $[ y>0 ]$ denotes Iverson bracket.

From here the distribution of $(M_t, M_t-B_t)$ is easy to read off: $$ f_{(M_t, M_t-B_t)}(x,y) = \sqrt{\frac{2}{\pi}} \frac{x+y}{t^{3/2}} \exp\left(-\frac{(x+y)^2}{2t} \right) [ y>0, x > 0 ] $$ Now finding the conditional probability density is also straightforward. Assuming $y>0$: $$ f_{M_t|M_t-B_t}(x|y) = \frac{f_{M_t,M_t-B_t}(x,y)}{f_{M_t-B_t}(y)}= \frac{x+y}{t} \exp\left(-\frac{x(x+2y)}{t} \right) [x >0] $$ This will now allow you to find the quantity of interest: $$ \mathbb{P}\left(M_t > a |B_t=M_t\right) = \mathbb{P}\left(M_t > a |M_t - B_t=0\right) = \lim_{y \to 0^+} \int_{a}^\infty f_{M_t|M_t-B_t}(x,y) \mathrm{d}x = \lim_{y \to 0^+} \exp\left(-\frac{a(a+2y)}{2t} \right) = \mathrm{e}^{-\frac{a^2}{2t}} $$