Order of dipole moments
I wondered about that for a long time, but Ben Norris and BigGenius have the truth of it: the question is about the bond dipole moment, otherwise the answer would make sense. Here's why.
Molecular dipoles

Sulfur trioxide has planar trigonal geometry, and its molecular dipole is zero. Similarly, molecular $\ce{SiO2}$ has the same geometry as $\ce{CO2}$, i.e. linear, and thus its molecular dipole is zero.

$\ce{Cl2O7}$ is bent at its central oxygen atom:

and thus has a non-zero dipole moment. While $\ce{P2O5}$ generally occurs as its “dimer” $\ce{P4O10}$, it is possible to write a molecular structure for $\ce{P2O5}$ which then strongly resembles that of $\ce{Cl2O7}$, being bent at the central O atom. It thus also has non-zero molecular dipole.

Finally, ordering between $\ce{Cl2O7}$ and $\ce{P2O5}$: because the P is less electronegative than Cl, the partial charges borne by the P atoms will be larger than those of the Cl atoms, and thus I expect the molecular dipole of $\ce{P2O5}$ to be greater than that of $\ce{Cl2O7}$.



Bond dipoles
If the question is talking about the $\ce{X–O}$ (or $\ce{X=O}$) bond dipole in the series where X = S, Si, P or Cl, it becomes easier to understand. For one thing, because you're asked about $\ce{X–O}$ bonds where X spans a row of the periodic table, which is typical of this sort of problems. The answer would then be that bond dipole moment get bigger when the electronegativity difference of X and O becomes larger (the sharing of electrons gets more asymmetric). This leads to the answer you gave.