Let a function $$f:\left[-\pi, \pi\right] \rightarrow $$ be expressed as $$f(x) : \sum_{j=1}^k a_{j} \cos(jx) + \sin x + 2 \sin 3x + 4 \sin 8x$$. f(-x) = -f(x), then

We want the function to be odd i.e, f(-x) = - f(x)
Now we know that sin(-x) = - sin (x) as sine is an odd function 
But cos (-x) = cos (x) as cosine is an even function

So if we put the input as -x in our defined function, we get, 
= $$\sum_{j=1}^k a_{j} \cos(-jx) + \sin (-x) + 2 \sin (-3x) + 4 \sin (-8x)$$
= $$\sum_{j=1}^k a_{j} \cos(jx) - \sin x - 2 \sin 3x - 4 \sin 8x$$
I we want both of these terms to be equal, we would have to remove the cosine term 
Option B does just that, by putting the coefficient of the cosine function as zero for all values of j. And the range given in option B is the same as the range of j in the given function.
This would give us the value of function to be  $$\sin (x) + 2 \sin (3x) + 4 \sin (8x)$$, which is an odd function. 
So option B is the correct answer.