Karp hardness of a guarding set in digraph
It is NP-complete (it is obviously in NP) by reduction from the dominating set problem.

Given an instance $G=(V,E)$ of the dominating set problem, split each vertex $v$ into two vertices $v_{\mathrm{in}}$ and $v_{\mathrm{out}}$, and for each edge $(u,v)\in E$, construct two edges $(u_{\mathrm{out}},v_{\mathrm{in}})$ and $(v_{\mathrm{out}},u_{\mathrm{in}})$. We claim that $G$ has a dominating set of size $k$ iff the newly-constructed graph (say $G'$) has a guarding set of size $2k$.

If $G$ has a dominating set $S$ of size $k$, easy to see $\{v_{\mathrm{in}}\mid v\in S\}\cup\{v_{\mathrm{out}}\mid v\in S\}$ is a guarding set of $G'$, which has size $2k$.

On the other hand, suppose $G'$ has a guarding set $S'$ of size $2k$. Let $S'_{\mathrm{in}}=S'\cap\{v_{\mathrm{in}}| v\in V\}$ and $S'_{\mathrm{out}}=S'\cap\{v_{\mathrm{out}}\mid v\in V\}$, then either $S'_{\mathrm{in}}$ or $S'_{\mathrm{out}}$ must has size at most $k$. Without loss of generality, say $\left|S'_{\mathrm{in}}\right|\le k$, then $S=\left\{v\mid v_{\mathrm{in}}\in S_{\mathrm{in}}'\right\}$ is a dominating set of $G$. This means $G$ has a dominating set of size $k$.