Multiply the <Karush-Kuhn-Tucker conditions> on the right by $\widehat\Omega$ and take the <matrix trace>:
$$
-p+\operatorname{Tr}(S\widehat\Omega)+\lambda\operatorname{Tr}(Z\widehat\Omega)=0.
$$
Symmetry and the defining property of the <subgradient of the absolute value> give
$$
\operatorname{Tr}(Z\widehat\Omega)
=\sum_{i,j}Z_{ij}\widehat\Omega_{ij}
=\sum_{i,j}|\widehat\Omega_{ij}|.
$$
Consequently the last two terms in the objective sum to $p$, and hence
$$
Q(\widehat\Omega)=-\log\det\widehat\Omega+p.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 205 / 2 / c / Solution

The <absolute value function> has
$$
\partial|x|=
\begin{cases}
\{\operatorname{sgn}x\},&x\ne0,\\
[-1,1],&x=0.
\end{cases}
$$

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 Incoming links: Subgradient of the absolute value