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**Hint:**To answer this question, we need to use the basic formula for the work done. We have to find the angle between the weight and the displacement of the suitcase. Substituting the value of the angle in the formula, we will get the required sign of the work done.

**Complete step-by-step solution:**

We know that while walking up on the steps, we cover distance in both the horizontal and the vertical directions. So the porter with the suitcase on his head will cover both the vertical and the horizontal distances. So along with moving forwards, he will also move vertically downwards. Since the suitcase is carried by the porter, the suitcase will also cover distances in the horizontal and the vertically upward directions.

Now, we know that the weight of any object acts vertically downwards. So the weight of the suitcase will act vertically downwards.

For the horizontal motion of the suitcase, the weight is perpendicular to the displacement. This means that ${{\theta }} = {90^ \circ }$. So the work done by the weight of suitcase becomes

${W_1} = Fd\cos {90^ \circ }$

$ \Rightarrow {W_1} = 0$

Now, for the vertically upward motion of the suitcase, the weight is opposite to the displacement. This means that ${{\theta }} = {180^ \circ }$. So the work done by the weight of suitcase becomes

${W_2} = Fd\cos {180^ \circ }$

$ \Rightarrow {W_2} = - Fd$

Therefore the total work done is given by

\[W = {W_1} + {W_2}\]

\[ \Rightarrow W = 0 - Fd = - Fd\]

Thus, the work done by the weight of the suitcase on the suitcase is negative.

**Hence, the correct answer is option B.**

**Note:**We only needed the sign of the work done, not its magnitude. So we didn’t worry about the magnitude of the horizontal and the vertical displacements covered by the suitcase.

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