Does rate of Evaporation change with surface area?
Why does vapour pressure decreases with decrease in surface area.  


Well, it does not. Vapour pressure is a function of temperature only. It is in fact the equilibrium constant for the reaction $\ce{H2O_{(l)}-&gt;H2O_{(g)}}$, and like all other equilibrium constants, is independent of surface area, amount of the species but depends only on the temperature. At a given T, only for one specific value of the partial pressure is water and its vapour in equilibrium.    

That said, water is not actually in equilibrium when in an open vessel evaporating into the atmosphere. The vapour pressure above the water filled vessel is far less than what is required for equilibrium. Water constantly evaporates as a desperate attempt to increase the vapour pressure to its equilibrium value, but since the vessel is open and the vapours not confined, the equilibrium value is seldom reached. If the surface area is larger, it means that the liquid phase can evaporate a larger amount of water while attempting to attain equilibrium. Therefore, larger surface area increases the amount of water that is lost as vapour while attaining equilibrium, since it is that part of the liquid exposed to the vapour phase which tries to maintain equilibrium by evaporating or condensing. Evaporation is essentially a surface phenomenon, as opposed to boiling which becomes considerably a bulk phenomenon.   

Since it is practically the surface layer which participates actively in the phase equilibrium, larger the surface, greater the attempt to attain equilibrium and quicker is the evaporation.