Nitriles react with Grignard reagents to form ketones, so the carbonyl carbon on each product must have came from a nitrile.
You also have the restriction that the longest carbon chain building must be three carbons.
a) Working backwards, the left side must have come from a nitrile, so the left side came from a Grignard.
How can we make each starting from propene?
To make the Grignard, prepare propyl bromide from propene via anti-Markovnikov HBr addition (HBr/peroxides), and then add Mg0/ether.
To make the nitrile, add sodium cyanide (NaCN) to propyl bromide.
Then add the nitrile and Grignard together, which forms the imine intermediate, which will hydrolyze to the ketone after you add water (H3O+).
b) This problem is harder because you're limited to starting materials with three carbons or less, so we have to make three "cuts" instead of two, like in a).
Working backwards again, prepare propyl Grignard from propene as described in a), and then add ethylene oxide to form 1-pentanol. Form the 6-carbon nitrile by first converting the alcohol to a bromide using PBr3, and then adding NaCN. Finally, add methyl Grignard followed by H3O+ to form the desired ketone.
