For $\theta>0$, the <Exponential martingale for Brownian motion> and the <Doob maximal inequality for a nonnegative submartingale> give
$$
\mathbb P\left(\sup_{t\leq s}B_t\geq a\right)
\leq\exp\left(-\theta a+\frac12\theta^2s\right).
$$
Taking $\theta=a/s$ gives $e^{-a^2/(2s)}$. Apply the same argument to $-B$ and use the <union bound>:
$$
\boxed{\mathbb P\left(\sup_{t\leq s}|B_t|\geq a\right)
\leq2e^{-a^2/(2s)}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 202 / 1 / 1 / 3 / Solution

{c}

For every $\theta\in\mathbb R$ and standard <Brownian motion> $B$,
$$
\exp\left(\theta B_t-\frac12\theta^2t\right)
$$
is a positive martingale. Applying <Itô formula> cancels the drift, and its expectation is one by the moment-generating function of the <normal distribution>.

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 Incoming links: Exponential martingale for Brownian motion