a+b+c=13 ---- 1
c+d+e=13 ---- 2
e+f+g=13 ---- 3
g+h+k=13 ---- 4
Add all four equations:
(a+b+c+d+e+f+g+h+k)+(c+e+g)=52
$\sum _{i=a} ^k i = 45$
c+e+g= 7 ---- 5
So, Option c and d are eliminated.
Now let's assume e=1, put in eq.2,3 and 5, add eq.2 and 3
c+d+2+f+g=26
c+d+f+g=24 ---- 6
c+1+g=7 from eq5
c+g=6 put in eq. 6
d+f=18, which is not possible
So, option a is eliminated.
Hence, Answer is B.