No. The <Omega combinator> is
$$
\Omega=(\lambda x.xx)(\lambda x.xx).
$$
Its only <beta-redex> contracts back to $\Omega$ itself. Every reduction sequence therefore repeats the same term, which is not in <beta-normal form>. Thus $\Omega$ has no beta-normal form.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 120 / 2 / e / Solution

{c}
{title2=$\Omega$}

The omega combinator is the divergent <untyped lambda calculus> term
$$
\Omega=(\lambda x.xx)(\lambda x.xx),
$$
whose only <beta reduction> reproduces $\Omega$.

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