H NMR question

All H's on the alpha carbon are equivalent because the carbon can rotate around its single bond to the carbonyl carbon. These therefore present as one peak with a relative integrated value of three. The carbonyl carbon lacks H's, so no signals there. The nitrogen in the amide group has two H's and a lone pair, but because the lone pair resonates with the carbonyl to exhibit partial double bond character, the nitrogen is by convention classified as sp2. Being sp2, the nitrogen and it's attached H's are trigonal planar, and that means one of the H's must be closer in space to the carbonyl oxygen, and one farther away. The H's therefore experience different local electronic environments and different shielding from the NMR magnetic field, resulting in distinct chemical shift values in the spectrum.

One signal from the 3 H's on the alpha carbon plus two signals from the two uniquely oriented H's on the amide equals three total signals.