By Gelfand representation of (real) C$^*$-algebras, it is known that $L_\infty[0,1]$ is isometrically isomorphic to $C(K)$, for some compact Hausdorff $K$. By looking at the proof, $K$ is actually defined in the following way: $$K=\{\mu\in L_\infty^*:\|\mu\|=\mu(e)=1\text{ and }\mu(fg)=\mu(f)\mu(g)\text{ for all }f,g\in L_\infty\},$$ where $e$ is the unit of the C$^*$-algebra (that is, in this case the 1-constant function). Moreover, it is proved that $K$ is non-empty since it contains the extreme points of the set $\{\mu\in L_\infty^*:\|\mu\|=\mu(e)=1\}$. Can we say anything about the cardinality of this set $K$?
So far I've been able to show the following topological lemma:
$\textbf{Lemma:}$ A compact Hausdorff countable space is metrizable.
Since $C(K)=L_\infty$ is non-separable, it is known that this implies that $K$ must be non-metrizable. This, combined with the lemma, shows that $K$ is uncountable. Can we do better than this?
The space $K$ you are interested in arises as the Stone space of the Boolean algebra of measurable subsets of the interval modulo the null sets, which is complete. All you have to do is to count the ultrafilters and for this, you can use the Balcar-Franek theorem to conclude there are precisely $2^{\mathfrak{c}}$ of them. Consequently,
$$|K| = 2^{\mathfrak{c}},$$
where $\mathfrak{c}$ is the cardinality of the contiuum.
A more general result may be found here (Proposition 2.15).