Step-by-step Analysis:

The question is about predicting the multiplicity (or splitting pattern) of protons in ¹H NMR for the given compounds. In NMR spectroscopy, the splitting pattern of a proton is determined by the number of neighboring protons (n) using the n+1 rule.

1. For Compound I:
   • The circled hydrogen is adjacent to two non-equivalent protons (one is another hydrogen in the ring and one is phenyl hydrogen).
   • This leads to a doublet of doublets (dd) splitting, due to coupling with two distinct sets of protons.
2. For Compound II:
   • The circled hydrogen has three different neighboring hydrogens due to different chemical environments (one from the methyl group and two different from the asymmetric carbon).
   • This results in a doublet of doublets of doublets (ddd) pattern.
3. For Compound III:
   • The circled hydrogen is adjacent to two equivalent hydrogens and experiences long-range coupling with an ethynyl hydrogen.
   • This creates a triplet of doublets (td) pattern due to two equivalent and one nonequivalent proton.

Conclusion: Based on the above analysis, the correct multiplicity assignments are:

Option: I = dd, II = ddd, III = td

This matches with the given correct answer.