[cseil]

Hello,
if I have an asymmetric ketone and it reacts with a secondary amine, I'll obtain the enamine.
But what enamine?

I'd say the second one because the double bond is more stable with more substituents. But my teacher once wrote as the product the first one and continued the reaction with other steps. I didn't realize at the time so I didn't ask her.

Another thing:
if I use LDA I'll deprotonate the less substituted carbon (LDA is a bulky molecule), right? What if I use another little base as KH?

[pgk]

You will get a mixture of both.
According your opinion which enamine will be predominant in the mixture and why?
If the methylcyclohexane is chiral, will it be retention or inversion of the chirality or a mixture of both stereoisomers and why?
If you draw the conformers (accoding streochemistry rules), you can access to the answer of the second question.

[cseil]

I'd say that the second product will be predominant. The double bond is more stable when there are more substituents.
EDIT: I'll think about the stereochemistry a little bit more because now I don't have enough time! I'll write it later 

Ps.
It's right if I say that if I want to deprotonate the less-substituted carbon I have to use LDA and if I want to deprotonate the more-substituted carbon I have to use another base as KH?

[pgk]

Yes, but the equilibrium is dynamic and not static. Remember, that there is a tautomerism between the enamines, as well between their chiral forms. The stronger the base, the faster the tautomerism is.
For your own education and for the same steric hindrance reasons, deprotonation of the more-substituted carbon is more convention via formation of the trimethylsilylenol ether, instead of using the enamine (Please,take a look to the literature, when you have time).