Major products A and B formed in the following , are

The given reaction sequence starts with 2-methylcyclohexanol:

Step 1: Formation of Product A
When 2-methylcyclohexanol is treated with phosphorus tribromide (${\mathrm{PBr}}_3$), a nucleophilic substitution reaction occurs. The hydroxyl group ($-\mathrm{OH}$) on the cyclohexane ring is replaced by a bromine atom ($-\mathrm{Br}$).
Thus, the major product A is 1-bromo-2-methylcyclohexane:
$\text{2-methylcyclohexanol}+{\mathrm{PBr}}_3\to \text{1-bromo-2-methylcyclohexane (A)}$

Step 2: Formation of Product B
When 1-bromo-2-methylcyclohexane (Product A) is treated with alcoholic KOH under heating (alc. KOH / Δ), it undergoes dehydrohalogenation via an E2 elimination mechanism.
The base abstracts a proton ($H^{+}$) from one of the adjacent $\beta$-carbons relative to the carbon bearing the bromine atom:
1. Abstraction of a proton from the tertiary $\beta$-carbon (C2, which holds the methyl group) yields a trisubstituted alkene, 1-methylcyclohexene.
2. Abstraction of a proton from the secondary $\beta$-carbon (C6) yields a disubstituted alkene, 3-methylcyclohexene.
According to Zaitsev's rule, the elimination reaction preferentially forms the more highly substituted, more stable alkene as the major product. Since a trisubstituted alkene is more stable than a disubstituted alkene, 1-methylcyclohexene is the major product.
Thus, the major product B is 1-methylcyclohexene.

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