Option A
R can have a multi-attribute key
C is correct
Relation not in 1NF if it has composite or multivalued attributes. (Thanks for correction @Philosophical_Virus)
D and B are ofcourse trivially wrong, take an example of any where we decompose any relation using 3NF decomposition, not after decomposition if it is in 3NF then for sure it will be in 1NF, having foreign key and more than one candidate key.