New node to be inserted is $P.$
Enqueue(){
P->Data=Data
P->Next=Head
Head=P
}
Time Complexity $=O(1)$ Because only pointer manipulation is involved which takes constant time.
Delete Tail
Dequeue(){
temp=head
While(temp->Next->Next!=NULL)
temp=temp->next
temp->next=NULL
tail=temp
}
Time Complexity = Time for Traversing list, free the last node and keep track of Tail pointer $= O(n)$
Answer is $B$