G:
S --> bSb|AcA
Ab --> A, Ab --> b
bA --> b ,
now start with S --> bSb
--> b (bSb) b
--> bb(AcA)bb [ we have only one choice left of c , that is bA-->b , on right we have two choice Ab --> A, Ab --> b ]
--> bbAc (Ab)b --> bbAc bb --> b (bA)cbb --> bb c bb
S=> bb c bb
again , bb(AcA)bb --> b(bA) c (Ab) b --> bb c Ab --> bb c b
No of b's on the right may be less.
so here L(G) = {bm c bn | m ≥ n, m,n ≥1} that leans L(G) is Context Free . But not regular.
Now you can see among those options only option B is matching.
now try to find L(G') also ..
for G'
S --> bSb|AcA
Ab --> A, Ab --> b
bA --> b , bA --> A
S --> b(S)b --> b bSb b --> b b b(S) b b b --> bb (b A) cA bbb
we have 2 choice here , either bA --> b , or bA --> A
by putting bA-->b we get S --> bbb c Abbb --> bbb c (Ab) bb --> bbb c bbb [ both left and right of c has 3 b's]
by putting bA--> A we get S --> bb A cAbbb --> b (bA) c (Ab)bb --> bb c Abb --> bb c bb [ both left and right of c has 2 b's ]
number of b can be vary in both sides of c .
so L(G') = { bm c bn | m,n ≥ 1 }
that means L(G') is regular .
Hence only option B is correct .
