The two answers come from two different questions, and the trick is to notice that only the second names a closed path.

The closed loop first — c → e → f → d → c. The final letter returns to the starting point, so this is a complete circuit and Kirchhoff's voltage law applies without qualification:

\(\sum_{\text{closed loop}}V=0\)

The sum of the rises and drops around any closed path in any circuit is always zero — no arithmetic is needed, and no knowledge of the element values. That disposes of one half of every option.

The partial path — c → e → f → d. This one stops at d without returning to c, so it is not a loop at all but a journey between two points. Its total must therefore equal the potential difference between the endpoints:

\(V_{cefd}=V_{c}-V_{d}\)

And \(V_{c}-V_{d}\) is simply the voltage across the element joining c and d directly — the 24 V across R₃.

So the partial path gives 24 V and the closed loop gives 0 V — option 1.

Why the two must be consistent. Completing the partial path by adding the final step d → c contributes \(V_{d}-V_{c}=-24\ \text{V}\), and

\(24+(-24)=0\ \checkmark\)

which is exactly KVL. That also shows why option 4's −24 V is merely the same path traversed in the opposite direction, c → d rather than d → c — the sign depends entirely on the direction chosen, so the labelling in the figure decides it.

The physical content of KVL is conservation of energy: carrying a unit charge around a closed path and returning it to its starting point must involve no net work, since the electrostatic potential is a single-valued function of position. Between two different points, by contrast, the work done is generally non-zero and is precisely the potential difference — and it does not matter which route is taken, which is why the long way round through e and f gives the same 24 V as the direct branch.

Hence, the partial loop gives 24 V and the complete loop gives 0 V.