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The logical adress =p1+p2+d.

Here p1=12bits p2=12bits and d=12 bits. Therefore logical adress size=36 bits
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logical address
no. of pages P page size

now no. of pages are P of logical address is further used for 1st level table...as

P
  P1( no. of pages of page table)12bit  P2(page size of page tableIi.e 4KW)12 bit

  no . of frames = p.a.s/ frame size
frame size =  p.a.s/ no. of frames 
frame size = 128MW/ 214 = 213 W
 now frame size same as page size...
page size = 213 W

so P = P1 +P2

AND  page size of logical address is same as frame size

logical addess
P1+P2= 12+12= 24 BIT  PAGE SIZE = FRAME SIZE==>13 BIT                         

LOGICAL ADDRESS = 37 BIT

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Please see if this approach is right or wrong.

Let Logical Address Space be x words.

Now, Page Size =  Frame Size = $\frac{Physical Address Space}{No. of Frames}$ = $\frac{2^{27}}{2^{14}}$ = 213 words.

Number of Pages = $\frac{x}{2^{13}}$

Page Table Entry = Number of bits to address each frame = 14bits = 1.75B

Number of pages in inner Page Table = $\frac{x * 1.75}{2^{13} * 2^{12}}$ (Each page of a Page Table is 4KW)

Now, $\frac{x * 1.75}{2^{13} * 2^{12}}$ = 212 (Page Table is divided into 212 Pages)

Solving, we get, x = 1.14 * 236 $\approx$ 21 * 36 = 237 = 37bits.

Please check this!!

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PAS = $2^{27}$W

with $2^{14}$ Frames

$\Rightarrow $ size of frame$ = $ size of Page $ =2^{13}$W

LAS =$13+12+12=37$ bits

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