How can the stereochemistry of the product of an E1cb reaction be determined?
Since your example compound has a chirality center, let's start by picking one enantiomer. We'll do the other enantiomer later. In the presence of a suitable base, this compound can be deprotonated to form the enolate anion. The enolate group itself can be E or Z, but for this purpose it does not matter.



In the second step of the mechanism, the electrons in the new pi bond from the enolate pi electrons shift toward the beta carbon. The new pi bond will be formed from the enolate p orbital on the alpha carbon and the C-Cl sigma-antibonding orbital on the beta carbon. 



For this to work, the C-Cl must be oriented perpendicular to the plane of the pi bond. We can draw Newman projections for our specific enantiomer of the enolate anion. There are two conformations that put the chlorine in the correct orientation. The second one below has less steric hindrance with the larger group (ethyl) near the H atom and the smaller group (methyl) oriented toward the enolate carbon. 



The major product forms from this conformation, with the methyl group cis to the enolate/carbonyl carbon. 



Now, the other enantiomer forms the same major product, but I would like you to convince yourself of that.