Ans-->3
It is always better to solve these types of question by a test case.
Bijection means Function which is both 1 to 1 as well as onto that is each element of the domain must map with a single element of the co-domain.
Say we have a set A = {1,2,3,4} & set B ={W,X,Y,Z} and the mapping be {1-> W, 2-> X, 3->Y,4->Z }
let E = {1,2} and F={2,3}
Now , f{EUF} = f{1,2,3} = {w,x,y} and similarly f{E} U f{F} = {W,X} U {X,Y} = {W,X,Y}
f{E ∩ F} = f{2} = {x} and similarly f{E} ∩ f{F} = {W,X} ∩ {X,Y} = X
So both are true
PS--> if it was given only a function but not a bijection then Q would have never been true..! Then only P should be true,this can be proved in the same way as above
Refer--> https://gateoverflow.in/721/gate2001-2-3.