This is $NP$-complete by tweaking Yuval's reduction.
We reduce from E3SAT like his reduction with a small modification that we are allowed to leave at most one clause unsatisfied.
So, for each edge $(C_i,l)(C_j,l')$ where $i<j \land l\neq\lnot l$, we have an arc $((C_i,l),(C_j,l'))$ (i.e. originates from $(C_i,l)$, ends in $(C_j,l')$). In other word, we orient each edge according to its endpoints clauses $C_i$ and $C_j$ such that $i<j$.
Now, for every pair of vertices $u=(C_i,l)$, $v=(C_j,l')$ in the undirected graph, if $dist(u,v)=2$ then we have $dist(u,v)=2$ also in our digraph, but not necessarily $dist(v,u)=2$.
To guarantee that each equidistant set in the undirected graph is also an equidistant set in our being constructed digraph, for each pair of vertices $u,v$ that $dist(u,v)=2$, we create a new verex $x_{uv}$ and add 2 arcs $(v,x_{uv})$, $(x_{uv},u)$. This guarantees that $dist(v,u)=2$.
It is also necessary to manipulate the case of having some $x_{uv}$ vertex in an equidistant set. But, this is impossible (as long as $k=m>4$ like in Yuval's reduction). To see this, suppose that $x_{uv}$ is included in some equidistant set. Clearly, from $x_{uv}$ going out, in one step, you must end at $u$. And, to reach $x_{uv}$ in one step, you must go from $v$. So, all the vertices that can be reached from $x_{uv}$ in two steps must have the form $(C_k,l'')$ where $i<k<j$ and $l''\neq\lnot l$ and $l''\neq \lnot l'$. That means we cannot have any other $x_{uv}$ vertex in our set. So, we can still satisfy at least $m-1$ clauses.
Conversely, if we have an assignment satisfying $m-1$ clauses, then in the corresponding vertex set, we have $u=(C_i,.)$ and $v=(C_j,.)$ where $C_i$ is the smallest-indexed satisfied clause and $C_j$ is the largest-indexed satisfied clause. Then add $x_{uv}$ to this set to form an $m$-equidistant set.