"All the problems in NP can also be reduced to problem X," means $X$ is NP-Hard (at least as hard as the hardest problem in NP)
$X$ is reducible to problem $Y$ which makes $Y$ also NP-Hard.
We are not sure if either $X$ or $Y$ is NP-Complete but if $Y$ is NP-Complete $X$ must also be NP-Complete as $X$ can be reducible to Y.
S1 : $X$ is NP complete and $Y$ is in NP ,
False. $X$ need not be NP-Complete as it can lie outside NP-class; same for $Y$.
S2 : $X$ is NP complete and $Y$ is NP hard
False, $X$ need not be NPC.
S3 : $X$ and $Y$ are NP complete and hence NP-hard
False. As both can lie outside NP-class making them NP-hard but not NP-complete.
S4 : $X$ and $Y$ are NP hard.
True, $X$ and $Y$ are definitely NP-hard as per the statements given.
So, S4 is the only correct statement, which is option A .