For what it's worth, here is a reduction showing NP-hardness. As regards $\Delta_2^p$-completeness, I'm not too optimistic, since one complete problem for this class is "given a formula, does its lexicographically maximal satisfying assignment set the last variable to true?" which seems harder than this problem.
We reduce from 3SAT. Let a formula $\phi'$ with $m-1$ clauses be given; add a fresh variable $x$ and add the one-literal clause $x$; this forces $x$ to be true. Thus we obtain a formula $\phi$ with $m$ clauses which is satisfiable if and only if $\phi'$ is.
Next we construct the graph $G$, which up to an extra component is identical to the graph used in Karp's original reduction showing that CLIQUE is NP-hard. Take a vertex $\ell$ for each literal occurrence in $\phi$. Include an edge between literal occurrences $\ell_1$ and $\ell_2$ if and only if (a) the occurrences are in different clauses, and (b) $\ell_1$ and $\ell_2$ are not contradictory (i.e. are negations of each other). Further, add a disjoint copy of the complete graph $K_m$ of $m$ vertices. This completes the description of $G$. As special vertex $v$ use the single literal occurrence of $x$.
Suppose the formula is satisfiable. Note first that there cannot be a clique among the literals of size larger than $m$ since every clique can contain at most 1 vertex per clause. Of course, among $K_m$, there is also no clique larger than $m$. Now take a satisfying assignment of $\phi$, and pick one true literal per clause. Since these literals cannot be contradictory, this forms a clique of size $m$, and includes the vertex $v$; it is of maximum cardinality $m$.
Suppose there is a maximum-cardinality clique in $G$ containing $v$. Then from the presence of $K_m$ in $G$, this clique must have size at least $m$. From what we said before, it can't have size larger than $m$, so it is of size exactly $m$. The clique thus chooses one literal per clause, and we can set them all true in a contradiction-free assignment. Hence $\phi$ is satisfiable.