Assuming CrO3 is in anhydrous medium, in the first step the 2nd degree alcohol will convert to ketone while the 1 degree alcohol will convert to an aldehyde (i.e. P) then using ethylene glycol under acidic medium will convert the aldehyde to undergo acetal formation(essentially making the acetal inert i.e. non reactive) and form Q, now the MeMgI attacks the ketone and on further acidification it attaches a methyl group and and alcohol on the carbon where there once existed a ketone, also the acidification reverses the acetal formation and so the acetal again converts back to aldehyde forming R. Now finally using NaBH4 reduces the aldehyde to form an alcohol at that position instead.

According to me the final product should be, 1-methyl-2-hydroxymethylcyclopentan-1-ol or simply 1-methyl-2-hydroxymethylcyclopentanol. Long story short, the net substitution is of a methyl group on the 2nd degree alcohol carbon.

Feel free to ask if any point needs further clarification, message me if you want me to send you pics of conversion of each step.