Let the veloercity of the particle after collision be v
As the horizontal components before and after collision remain unchanged then
ucosQ = v cos(90-Q)=vsinQ. ....(1)
Again the coefficint of restitution
e=(vsin(90-Q))/(usinQ)=3/4 .......(2)
Comparing (1) and (2) we get
(v/u)"2=3/4
=>v= (√3/2)u
Option (1) is correct