The way NEW Edge Weights are defined, for any path $Y$ from source $s$ to destination $d$ in $G$, the new length will become:
$\text{Old Length} + f(s) - f(d).$ So, the NEW length of any path ONLY depends on the source and the destination node of the path, NOT on any intermediate vertices on the path(intermediate vertices $f(v)$ values will cancel out, Only source’s $f(s)$ and destination’s $f(d)$ will remain).
For any two vertices $s,d;$ assume that we have two paths $P,Q$ from $s$ to $d.$
Assume in the Old Graph, Length of $P = P_{OLD}$ and Length of $Q = Q_{OLD} $ and assume that $P_{OLD} \leq Q_{OLD}.$
In the NEW Graph, Length of $P = P_{NEW} = P_{OLD} + f(s) – f(d)$ and
Length of $Q = Q_{NEW} = Q_{OLD} + f(s) – f(d).$
NOW, Since $P_{OLD} \leq Q_{OLD},$ So, WHATEVER real values $f(s),f(d) $ have, we have $P_{NEW} \leq Q_{NEW}.$
So, this statement is Correct: For any mapping of vertices to real values, the shortest paths won't change BUT length of Shortest path will definitely change(by a value $f(s) – f(d)$).
Regarding Option $D:$
It doesn’t matter what is $f(u)$ for any $u.$ So, WHATEVER option is created regarding $f$ value of vertices, it will be only sufficient condition. Only option $A$ is necessary & sufficient condition.
Just a small “Irrelevant to the question” note that Option $D$ is Not necessarily making all $f(u) = 0.$ In case of negative weights, for some vertices $f$ can be negative in option D. BUT again WHATEVER $f$ we have, it doesn’t matter.