Here we have 11 nodes hence 11 positions. To satisfy topological ordering we have to fix node a at first position, node k at 11th position, and node d at 4th position.
Now we have 2,3,5,6,7,8,9,10 as empty positions. Among them we can fill positions 2 and 3 from nodes b or c only (hence 2 ways) and remaining 6 places can be filled by e,g,i,f,h,j such that relative ordering e-g-i and f-h-j must be maintained, this can be done in 6!/(3!3!) = 20 ways.
Total topological orderings = 2 x 20 = 40