Role of alcoholic potassium hydroxide in an E2 reaction
Haloalkanes react with alc. $\ce{KOH}$ mainly via $E_2$ mechanism. But, in some cases, where $\beta-\ce{H}$ isn't present, it proceeds via $E_1$, as in case of $(4)$ one. Let's analyze each option,
Option $(1)$:
Firstly, the $\ce{-Cl}$ departs, and then the base picks up a $\beta-\ce{H}$ from the terminal methyl group (i.e., $\ce{C-5}$), resulting in mono- substituted alkene (Hofmann product as major). Steric factors are significant in selectivity of $\beta-\ce{H}$. Therefore, Zaitsev product will be formed in very minor concentrations, bcoz $\ce{-C(CH3)2Ph}$ is more bulkier substituent than $\ce{^tBu}$.
Option $(3)$:
This one is similar to $(1)$ case, except for the fact that due to less steric hindrance, it will give di- substituted alkene (Zaitsev product).
Option $(2)$: It's clear that this molecule will form a tri- substituted alkene.
Option $(4)$: As hinted above, it will proceed through $E_1$, resulting in a tri- substituted alkene.

Different levels of substitution are shown below, (source: www.chemistrysteps.com)