Step 1: Link vapour pressure to the solvent fraction:
For a non-volatile solute in a dilute solution, the relative lowering $\frac{\Delta p}{p^{\circ}}$ is about equal to the mole fraction of dissolved particles. A larger number of dissolved particles gives a larger drop, so we look for the least number of particles.
Step 2: Count particles per mole:
Take $1$ mol of each solute in water. NaCl gives $2$ mol of ions. $\text{CaCl}_2$ gives $3$ mol. $\text{AlCl}_3$ gives $4$ mol, assuming complete dissociation. Sucrose stays as $1$ mol of molecules, because it has no ionic bond to break.
Step 3: Rank the vapour pressures:
Order of particle number: sucrose $(1) < \text{NaCl}\ (2) < \text{CaCl}_2\ (3) < \text{AlCl}_3\ (4)$. Vapour pressure runs the other way: sucrose has the highest and $\text{AlCl}_3$ the lowest. So options (A), (B) and (C) are all lower than option (D).
Final Answer:
Sucrose gives the fewest particles, so the highest vapour pressure is for option (D).
\[ \boxed{\text{Option (D)}} \]