Let $$P_n = \alpha^n + \beta^n$$, $$n \in \mathbb{N}$$. If $$P_{10} = 123$$, $$P_9 = 76$$, $$P_8 = 47$$ and $$P_1 = 1$$, then the quadratic equation having roots $$\frac{1}{\alpha}$$ and $$\frac{1}{\beta}$$ is:

We are given $$P_n = \alpha^n + \beta^n$$ with $$P_{10} = 123$$, $$P_9 = 76$$, $$P_8 = 47$$, and $$P_1 = 1$$.

If $$\alpha$$ and $$\beta$$ are roots of $$x^2 - sx + p = 0$$ where $$s = \alpha + \beta$$ and $$p = \alpha\beta$$, then by Newton's identity: $$P_n = s \cdot P_{n-1} - p \cdot P_{n-2}$$.

Using $$n = 10$$: $$P_{10} = s \cdot P_9 - p \cdot P_8$$, so $$123 = 76s - 47p$$ ...(1)

We also know $$P_1 = \alpha + \beta = s = 1$$.

Substituting $$s = 1$$ in (1): $$123 = 76 - 47p$$, so $$47p = 76 - 123 = -47$$, giving $$p = -1$$.

Verification: $$P_2 = s \cdot P_1 - 2p = 1 - (-2) = 3$$. $$P_3 = s \cdot P_2 - p \cdot P_1 = 3 + 1 = 4$$. Continuing this recurrence should yield $$P_8 = 47$$, $$P_9 = 76$$, $$P_{10} = 123$$. Let us verify: $$P_4 = P_3 + P_2 = 7$$, $$P_5 = 11$$, $$P_6 = 18$$, $$P_7 = 29$$, $$P_8 = 47$$, $$P_9 = 76$$, $$P_{10} = 123$$. Confirmed.

The quadratic with roots $$\alpha$$ and $$\beta$$ is $$x^2 - x - 1 = 0$$ (since $$s = 1$$ and $$p = -1$$).

We need the quadratic with roots $$\frac{1}{\alpha}$$ and $$\frac{1}{\beta}$$. The sum is $$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{1}{-1} = -1$$. The product is $$\frac{1}{\alpha\beta} = \frac{1}{-1} = -1$$.

So the quadratic is $$x^2 - (-1)x + (-1) = 0$$, which is $$x^2 + x - 1 = 0$$.

Hence, the correct answer is Option B.