The reaction you've provided is an example of an esterification reaction, where methanol (CH3OH) and acetic acid (CH3COOH) react to form methyl acetate (CH3COOCH3) and water (H2O), with a negative change in enthalpy (ΔH = -8.5 kJ/mol). Which is same same , CH3OH + CH3COOH ⇌ CH3COOCH3 + H2O.
I think by adding sodium hydroxide (NaOH) to the mixture can indeed shift the equilibrium of this reaction to the left. This is because sodium hydroxide is a strong base that can react with acetic acid (CH3COOH) to form the acetate ion (CH3COO-) and water. This reaction is a neutralization reaction,
CH3COOH + NaOH → CH3COO- Na+ + H2
By removing some of the acetic acid from the reaction mixture through this neutralization. Try to apply Le Chatelier's principle
Hope you have understood.