The decision version of this problem is NP-hard (and so NP-complete), by reduction from Hitting Set.
Let $S_1,\ldots,S_m$ be an instance of hitting set. We will have the following vertices and edges:
- Vertices $A_i$ for $1 \leq i \leq m$ and $A_{i,j}$ for $1 \leq i \neq j \leq m$.
- Vertices $B_u$ for every $u$ in the universe of the hitting set instance, and $B_{i,j}$ for $1 \leq i \neq j \leq m$.
- Vertices $C_i$ for $1 \leq i \leq m$ and $C_{i,j}$ for $1 \leq i \neq j \leq m$.
- For every $u \in S_i$, edges $A_i \to B_u \to C_i$.
- For every $1 \leq i \neq j \leq m$, edges $A_{i,j} \to B_{i,j} \to C_{i,j}$.
- For every $1 \leq i \neq j \leq m$, edges $A_i \to B_{i,j} \to C_j$.
Every feasible solution to this instance of your problem has to contain each $B_{i,j}$, since this is the only way to connect $A_{i,j}$ and $C_{i,j}$. Hence every $A_i$ is automatically connected to every $C_j$ for $j \neq i$. In order to connect $A_i$ to $C_i$, we need to have $B_u$ for some $u \in S_i$. Hence the minimal solution contains $m(m-1) + k$ vertices iff the hitting set instance has a minimal solution with $k$ elements.