Signal A corresponds to the proton in the double bond because it is a doublet of triplets,
coupling with B and E, with a coupling constant of ~16Hz across the double bond.
It is more deshielded because of the resonance structure that can be formed with the C=O.
This makes B the other proton in the double bond. Notice that they have the same JH.
C is the proton of the isopropyl group, and it is more deshielded than a normal sp3 proton would be because of the electron
withdrawing effects of the oxygens.
D corresponds to the protons of the methoxy group, which are singlets because the oxygen removes
the possibility of interactions with neighbors.
E is the two sp3 protons, and the ddd comes from interactions with B and C.
Finally, F is the three methyl groups off of the tert-butyl. They are all identical
and there is only one signal with 9 H, so this is a clear choice.