The Hoffman elimination first turns an amine into a better leaving group by methylating it with excess methly iodide, and then eliminates the methylated amine using base (such as -OH fomed from Ag2O in H2O). The products are alkenes.
This reaction is similar to an E2 elimination in that a beta proton is ripped off, and the alkene forms between alpha and beta. One notable difference from the E2 is that the less substituted product is favored, so Hoffman products are usually anti-Zaitsev.
a) There are three different types of beta protons, so three possible products can form. Product A is the least substituted alkene, so it's the major product.
b) There is only one possible product. The Hoffman elimination can't be done on aryl hydrogens (benzene rings).
