well, there is no net charge for Cl2, PF5, and BH3 which means that the only molecules left are NH3 and SO3.

The reason Cl2, PF5, and BH3 have no net charge is because each Cl molecule in Cl2 has the same electronegativity which cancels each other out. Similarily, the net charge of each P-F bond in PF5 cancels each other out because in its sp3d hybridization, PF5 is trigonal bipyramidal having 2 P-F bonds cancel out in one plane, and 3 P-F bonds cancel out in the other plane. Even though the B-H bonds in BH3 are polar, the polarity among each B-H bond is the same. Since each B-H bond is 120 degrees apart, the overall net charges expressed in x-axis and y-axis vector quantities are cancelled out to equal 0. Thus, BH3 has no net charge.

S and O have very close electronegativities so even though SO3 will have an overall net dipole charge, it will probably be so small that SO3 will not even be considered polar.

That just leaves NH3, which is in fact polar because the difference in electronegativities is vastly greater than that of SO3. In addition, you know that the charges in NH3 will not cancel out because NH3 has an sp3 hybridization.

Hybrization is the reason NH3 is polar but BH3 is not. BH3 has a hybridization of sp2 making it trigonal planar. NH3 with its sp3 hybrization is tetrahedral due to an extra non-bonding e- pair.

And that is why NH3 is polar 🙂