I think Answer is option (C).
I think we have to compare overhead of Two Level Paging, Segmentation and Segmented Paging.
Let us try to find find optimal answer for all three cases -
Case 1(Two Level Paging):
Assuming outer page tables are not page aligned. But inner one is page aligned.
For $P_{1}$
Size of entries in second level = 195*4 B = 780 B ~= 1KB
Size of entries in first level = 4 B
For $P_{2}$
Size of entries in second level = 254*4 B = 1016 B ~= 1KB
Size of entries in first level = 4 B
For $P_{3}$
Size of entries in second level = 45*4 B = 180 B ~= 1 KB ( Because 1 Page is enough)
Size of entries in first level = 4 B
For $P_{4}$
Size of entries in second level = 364*4 B = 1465 B ~= 2 KB
Size of entries in first level = 8 B
So P = 5*1024 + 20 = 5140 B
Case 2 (Segmentation):
Assuming segment table is not page aligned -
S = (4+5+3+8)*8 = 160 B
Case 3 (Segmented Paging):
Assuming page table is not page aligned.
For $P_{1}$
Size of segment table = 4*8 B
Size of entries in page table entries corresponding to all segment = 195*4 B = 780 B ~= 1KB (We can address in power of 2)
For $P_{2}$
Size of segment table = 5*8 B
Size of entries in page table entries corresponding to all segment = 254*4 B = 1016 B ~= 1KB
For $P_{3}$
Size of segment table = 3*8 B
Size of entries in page table entries corresponding to all segment = 45*4 B = 180 B ~= 0.25 KB
For $P_{4}$
Size of segment table = 8*8 B
Size of entries in page table entries corresponding to all segment = 364*4 B = 1465 B ~= 2 KB (We can address in power of 2)
So T = 160 + 4.25*1024 = 4512 B
Please correct if something is not proper.