Determination of major product in crossed aldol condensation
There are two main steps in any aldol condensation:-
1. Formation of enolate ion
2. Using the enolate ion as a nucleophile.

The 1st step is a reversible step. If you think that the kinetically favoured enolate ion will be formed fast then you are wrong as it also disappears faster. So, in any case the one which is more thermally stable will form in more amount except in the cases where the formation of stable enolate is too slow or the backward reaction of kinetic favoured enolate is blocked or not enough time has been given to let the reaction reach equilibrium.


Now, I will tell you about above strategies to block reverse reaction or make formation of stable enolate too slow:-

1. Use a strong electron withdrawing group at alpha position to the newly formed CH2(-)COR so that the electron can spend most of time away from the original carbon.
-This blocks reverse reaction

2. Use a bulky base or somehow hinder the alpha carbon to make the extraction of proton hard which will in turn make the reaction slow.

3. Using Non-Protic Solvent stops the protonation of the enolate ion which in turn blocks the reverse reaction.

Now, the answer for your question, the conjugate base of acetaldehyde is more stable when compared to the former but the attack of conjugate base of acetophenone is more easier as compared to latter because of steric reasons and the more electrophilic character of aldehyde and when you make cannonical structures, the number of structures which has negative charge on oxygen is greater than being away which makes it a good nucleophile when you take weighted average of all. So, the major product should be 1-phenyl-but2en1one.