Segment Address : Segment No(S)+WordNo(D)
Since Segment is divided into 8K pages of 2K words each.It means Paging is done over D part of address . Now D Part will be broken into P|D' Format.
Where P is PAge No
D'- Word No
So D=P+D'=log(8K)+log(2k)=13+11=24
Now Segment table is also divided.It means Paging is done on S part Of Logical address.
S=Log(256K)+Log(512)=18+9=27
So Logical Address=S+D=27+24=51 Bits