Glucose when heated with ${\mathrm{CH}}_3\mathrm{OH}$ in presence of dry $\mathrm{HCl}$ gas gives $\alpha$ and $\beta$ methyl glucoside because it contains

 Glucose reacts with methanol, in presence of dry $\mathrm{HCl}$, to form alpha and beta methyl glycosides.

${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6+{\mathrm{CH}}_3\mathrm{OH}\to \alpha -{\mathrm{C}}_6{\mathrm{H}}_{11}{\mathrm{O}}_6{\mathrm{CH}}_3+\beta -{\mathrm{C}}_6{\mathrm{H}}_{11}{\mathrm{O}}_6{\mathrm{CH}}_3$

The formation of alpha and beta methyl glycosides is due to the ring structure of glucose, as the methyl alcohol can attack the anomeric carbon in 2 ways, leading to these two configurations.

Hence the correct answer is (C) P-alcoholic group.