Karp hardness of a clique with given number of outer-incident edges
Our problem is $NP$-complete by a reduction from Exact Cover by $3$-sets ($\mathrm{X3C}$).

Given an $X3C$ instance with the universe set $U=\{e_1,e_2,\cdots,e_n\}$ and collection of $3$-sets $\mathcal{C}=\{s_1,s_2,\cdots,s_m\}$, we create an undirected graph $G$ as follows.

For each element $e_i$, create a vertex (which will be referred to by the same name).

For each $3$-set $s_j$, create a vertex (which will also be referred to by the same name).

Connect $e_i$ and $s_j$ whenever $e_i\in s_j$.

Connect $s_{j_1}$ and $s_{j_2}$ whenever they are disjoint.

Pick some large enough $M$, connect each set-vertex to dummy vertices to make sure that every set-vertex is of degree $M$.

Call the obtained graph $G$. Now, set $k=n+(M-3-\frac{n}3+1)*\frac{n}3$. The reduction then outputs an instance, namely $(G,k)$.

Suppose there exists an exact cover for the $X3C$ instance, we can easily form a solution to our problem by taking all the set-vertices of the exact cover. Clearly, this is a clique since every pair of $3$-sets in an exact cover needs to be disjoint. It has exactly $k$ outer-incident edges.

Conversely, if there exists a clique with $k$ outer-incident edges then every vertex of this clique needs to be set-vertex. Indeed, if there is any element-vertex in the clique, then the size of the clique is equal to the number of $3$-sets that includes that element plus one (the element-vertex itself). But then the size of this clique is only $4$, since $X3C$ is still $NP$-complete when each element is included in at most three $3$-sets. So, this is a clique of set-vertices. Thus, they are pairwise-disjoint. With $k$ outer-incident edges, the size of this clique must be exactly $\frac{n}3$. This is due to the fact that $M$ can be large enough (still polynomially) so that the binomial defining $k$ (in variable $x$, the size of the clique) is monotone in $[1,\cdots,m]$.