= Solution
Let $A=\langle M\rangle$ be the <quadratic variation> and stop at $T_b=\inf\{u\geq0:A_u>b\}$. Its continuity and monotonicity imply $A_{u\wedge T_b}\leq b$ and $A_{T_b}=b$ whenever $T_b<\infty$. The standard <martingale> isometry after stopping by bounded <quadratic variation> gives $\mathbb E(M_{t\wedge T_b}^2)=\mathbb EA_{t\wedge T_b}\leq b$; in particular the stopped <local martingale> is <square-integrable>. This criterion follows by localization and the <Doob L2 maximal inequality>, which bound the second <moment> of the stopped supremum by $4b$ and then permit removal of localization.
On $\{A_t\leq b\}$ one has $T_b\geq t$, so the original and stopped paths agree through time $t$. Therefore the <Doob L2 maximal inequality> and the <Markov inequality> give
$$
\begin{aligned}
\mathbb P\left(\sup_{s\leq t}M_s>a\right)
&\leq\mathbb P\left(\sup_{s\leq t}|M_{s\wedge T_b}|>a\right)+\mathbb P(A_t>b)\\
&\leq\frac{4b}{a^2}+\mathbb P(A_t>b).
\end{aligned}
$$
Thus
$$
\boxed{\mathbb P(\sup_{s\leq t}M_s>a)\leq4b/a^2+\mathbb P(\langle M\rangle_t>b).}
$$
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