Rationalizing enamine stabilities The solutions manual said that enamine B with its roughly sp2 methyl group can force the nitrogen out of the plane and this messes around with the conjugation. 

How about the fact there is a significant degree of resonance donation by the nitrogen, which places a significant partial negative charge on carbon. In enamine A, this carbon is secondary. In enamine B, this carbon is tertiary. Extra inductive donation from the methyl group destabilizes the product.
Conjugation effects are not very important in this case. 
The main reason is steric hindrance. If the more substituted alkene is formed, then both the methyl group and the nitrogen ring will be on the same plane, and repulsions will destabilize the molecule.

In the case of A, the methyl group is more free to move in a pseudo-chair conformation of the cyclohexene ring.

But there are exceptions, for example:



In the second case the more substituted alkene is formed, because is more stable and there is essentially no difference in regard to steric hindrance.