There's a hint:
We need to use
$$\Omega_{\varepsilon} = \left\{ x \in \Omega \ : \ |a(x)| > \|a\|_{\infty} - \varepsilon \right\}$$
which is of positive measure, and what we have to do is to write out $\|M_a u_{\varepsilon}\|$ where $u_{\varepsilon} = u \chi_{\Omega_{\varepsilon}}$
I don't really know what to do. If we write out this norm, we get eventually: $$\|M_a u_{\varepsilon}\| > (\|a\|_{\infty} - \varepsilon) \|u_{\varepsilon}\|$$
So what am I supposed to do? We can't do something like $$\|a\|_{\infty} \|u_{\varepsilon}\| \ge \|M_a u_{\varepsilon}\| > (\|a\|_{\infty} - \varepsilon) \|u_{\varepsilon}\|$$
and take $\varepsilon \rightarrow 0$, because we'll get a contradiction because of that "$>$" (and also, $\Omega_{\varepsilon} \rightarrow \varnothing$ because no value $|a(x)|$ is greater than the supremum norm).
The point is the other inequality is more obvious: $$ \|M_a u\|_p^p=\int_\Omega|a|^p|u|^p\leq \|a\|_\infty^p\|u\|_p^p. $$ Now, with the hint, for given $\epsilon>0$, you can find an element in $L^p$ (be careful with the possibility that $\mu(\Omega_\epsilon)=+\infty$ if $\Omega$ is not of finite measure) with $$\|M_au_\epsilon\|> (\|a\|_\infty-\epsilon)\|u_\epsilon\|$$ so that $\|a\|_\infty$ is in fact the supremum in the definition of the operator norm.
Edit: Some more details follow. Fix $\epsilon>0$ and let $u_\epsilon=\chi_{\Omega_\epsilon}$ and note that $u_\epsilon$ is nonzero as an element in $L^p$ since $\mu(\Omega_\epsilon)>0$. Then, by the definition of the operator norm as a supremum, $$ \|M_a\|\geq\frac{\|M_au_\epsilon\|}{\|u_\epsilon\|}> \|a\|_\infty-\epsilon. $$ Now, since $\epsilon>0$ was arbitrary, we can take $\epsilon\to 0$ in the inequality $$ \|M_a\|>\|a\|_\infty-\epsilon $$ and conclude $\| M_a\|\geq \|a\|_\infty$