 Solution

Use vertex splitting to encode node failures as well as link failures. Replace each vertex $v$ by $v_{\mathrm{in}},v_{\mathrm{out}}$ and an internal arc $v_{\mathrm{in}}\to v_{\mathrm{out}}$. Give that arc capacity one for an intermediary node, and capacity $B=|E|+|V|+1$ for the client and servers. Replace an undirected link $uv$ by arcs $u_{\mathrm{out}}\to v_{\mathrm{in}}$ and $v_{\mathrm{out}}\to u_{\mathrm{in}}$, each of capacity one. Connect server outputs to a common sink with capacity $B$ and take $c_{\mathrm{out}}$ as source.

A minimum cut avoids capacity-$B$ arcs because failure of all original links provides a cheaper separating set. We can normalize its source side so that $v_{\mathrm{out}}$ being on that side implies $v_{\mathrm{in}}$ is also there: moving $v_{\mathrm{in}}$ to that side cannot increase the cut capacity, since its sole outgoing arc leads to $v_{\mathrm{out}}$. In such a cut, at most one direction of any original link crosses. Every unit internal arc crossing corresponds to failing its intermediary node; every unit link arc corresponds to failing that link. Their failures disconnect all servers, so the cut capacity is the cost of a real failure set.

Conversely remove the internal arcs of failed nodes and both arcs of failed links. The reachable-side cut contains only arcs corresponding to those failures, with at most one crossing orientation per failed link; its capacity is at most their number. The max-flow min-cut theorem thus identifies the minimum mixed failure count ${\lambda }_{\mathrm{mix}}$ exactly. HenceThe split graph has $O(|V|)$ vertices and $O(|E|+|V|+|S|)$ arcs, so the same Edmonds–Karp algorithm gives a polynomial-time decision procedure.