I think option B is right answer, and i think there was a mistake in calculating the storage overhead for segmentation+paging portion. For rest calculations it is correct. So,P=9KB, S=4KB

First realize that in segmentation we divide a process into segments and we create page table for each segment of the process. Here as it is mentioned that 2 level paging has been used,so here it goes like this:

Note:- First see that max segment size can be=256KB so a segment can have max=256 pages, for 256 pages we need a page table of size=(256*4)=1024B=1KB=page size, so we need max 1 page to store the page table of a segment,but as it is 2 level paging we need atleast 2 pages i.e 2KB to store the page table of a segment.

for P1, no of segments=4,for each segment of the process we need a separate TWO LEVEL PAGING.so (4*2)KB=8KB for page table of P1 and for segment table we need another page,so another 1KB..so for P1 total 9KB

for P2, no of segments=5,for each segment of the process we need a separate TWO LEVEL PAGING.so (5*2)KB=10KB for page table of P2 and for segment table we need another page,so another 1KB..so for P2 total 11KB

for P3, no of segments=3,for each segment of the process we need a separate TWO LEVEL PAGING.so (3*2)KB=6KB for page table of P3 and for segment table we need another page,so another 1KB..so for P3 total 7KB

for P4, no of segments=8,for each segment of the process we need a separate TWO LEVEL PAGING.so (8*2)KB=16KB for page table of P4 and for segment table we need another page,so another 1KB..so for P4 total 17KB

So total, T=(9+11+7+17)KB=44KB

So, S<P<T option (B)