Major products A and B formed in the following , are

Based on the reaction sequence shown in the images, the step-by-step conversion is as follows:

Step 1: Formation of Product A
The starting material shown in the reaction scheme is 2-methylcyclohexanol. When treated with phosphorus tribromide (${\mathrm{PBr}}_3$), the hydroxyl group (-OH) is substituted by a bromine atom (-Br). This nucleophilic substitution reaction proceeds via an $S_{N}2$ mechanism, replacing the -OH group with -Br to yield 1-bromo-2-methylcyclohexane as the major product A (shown in the first option image):
$\text{2-methylcyclohexanol}+{\mathrm{PBr}}_3\to \text{1-bromo-2-methylcyclohexane (A)}$

Step 2: Formation of Product B
When 1-bromo-2-methylcyclohexane (Product A) is heated with alcoholic KOH ($\text{alc. KOH}/\Delta$), it undergoes dehydrohalogenation (elimination of HBr) via an E2 mechanism. Under these strongly basic and hot conditions, the hydroxide/ethoxide base abstracts a beta-proton from an adjacent carbon atom while the bromide leaving group departs.
There are two adjacent beta-carbons from which a hydrogen atom can be eliminated:
1. Elimination from the C-2 carbon (which holds the methyl group) yields 1-methylcyclohexene (a trisubstituted alkene).
2. Elimination from the C-6 carbon yields 3-methylcyclohexene (a disubstituted alkene).
According to Zaitsev's (Saytzeff's) rule, the major product of an elimination reaction is the more highly substituted, thermodynamically more stable alkene. Therefore, 1-methylcyclohexene is formed preferentially as the major product B:
$\text{1-bromo-2-methylcyclohexane (A)}+\text{alc. KOH}/\Delta \to \text{1-methylcyclohexene (B)}$

Thus, the structures corresponding to A and B are:
A = 1-bromo-2-methylcyclohexane
B = 1-methylcyclohexene