Now check this diagram, this is forest obtained from above given graph using Kruskal's algorithm for MST.

So according to the question edge $d-e$ has weight 5 and it is included in the formation of MST. Now if edges $b-e$ and $e-f$ has weight greater than 5 than it is not a problem for our MST because still we will get the given tree as Kruskal's algorithm takes the smallest weighted edge without forming a cycle.

Cost of edge $b-c \geq 4$ may also lead us to the same tree as above though Kruskal's algorithm will have choice between $c-f$ and $b-c$.

Now if the edge weight of $a-d$ becomes 4, it is guaranteed that Kruskal's algorithm will not select edge $d-e$ because its edge cost is 5, and hence the tree structure will change. But there can be the case where edge weight is greater than 4 and we still get the same tree (happens when $a-d \geq 5$). Because in the question they asked to point out an unnecessary condition this case is not the answer as we need $a-d \geq 5$ which implies $a-d \geq 4.$

Now notice option A. Put $a-b = 5.$ The given MST would not change. So, this condition is not always necessary and hence is the answer..

Therefore option A is the answer .