If three positive numbers $$a$$, $$b$$ and $$c$$ are in A.P. such that $$abc = 8$$, then the minimum possible value of $$b$$ is:

Let the three positive numbers be $$a$$, $$b$$ and $$c$$ with $$a<b<c$$; they are stated to be in arithmetic progression, so the common difference can be denoted by $$d\ge 0$$ and we may write

$$a=b-d,\qquad b=b,\qquad c=b+d.$$

The condition of the problem gives the product

$$abc=8.$$

Because the three numbers are positive, we may compare their arithmetic mean (AM) with their geometric mean (GM). Calculating the AM,

$$\text{AM}=\frac{a+b+c}{3}=\frac{(b-d)+b+(b+d)}{3}=\frac{3b}{3}=b.$$

Next, using the given product, the GM is

$$\text{GM}=(abc)^{1/3}=8^{1/3}=2.$$

A fundamental inequality for any three positive numbers is

$$\text{AM}\ge\text{GM}.$$

Substituting the values we have just found,

$$b\ge 2.$$

This inequality shows that $$b$$ cannot be smaller than $$2$$. To see that $$b=2$$ is actually attainable, we choose $$d=0$$, giving

$$a=b=c=2,$$

and then

$$abc=2\cdot 2\cdot 2=8,$$

exactly as required. Thus $$b=2$$ is not only a lower bound but is also achievable, making it the minimum possible value.

Hence, the correct answer is Option B.