The correct option is D A=Depolarised, B=Polarised
At site A: When a stimulus is applied at a site at the membrane it becomes freely permeable to Na+. This leads to a rapid influx of Na+ followed by the reversal of the polarity at that site i.e., the outer surface of the membrane becomes negatively charged and the inner side becomes positively charged. The polarity of the membrane at the site is thus reversed from the resting potential and hence it is called depolarised.
At site B: The impulse has not reached, it is in resting phase. The axonal membrane is more permeable to potassium ions (K+) and nearly impermeable to sodium ions (Na+) and to negatively charged proteins present in the axoplasm. As a result, at site B the outer surface of the axonal membrane possesses a positive charge while its inner surface becomes negatively charged and therefore is polarised.