If the coefficient of $$x^7$$ in $$\left(ax - \frac{1}{bx^2}\right)^{13}$$ and the coefficient of $$x^{-5}$$ in $$\left(ax + \frac{1}{bx^2}\right)^{13}$$ are equal, then $$a^4 b^4$$ is equal to:

In $$\left(ax - \frac{1}{bx^2}\right)^{13}$$, the general term is:

$$T_{r+1} = \binom{13}{r}(ax)^{13-r}\left(\frac{-1}{bx^2}\right)^r = \binom{13}{r}a^{13-r}\frac{(-1)^r}{b^r}x^{13-3r}$$

For $$x^7$$: $$13 - 3r = 7 \Rightarrow r = 2$$

Coefficient = $$\binom{13}{2}a^{11}\frac{1}{b^2} = 78\frac{a^{11}}{b^2}$$

In $$\left(ax + \frac{1}{bx^2}\right)^{13}$$, the general term is:

$$T_{r+1} = \binom{13}{r}(ax)^{13-r}\left(\frac{1}{bx^2}\right)^r = \binom{13}{r}\frac{a^{13-r}}{b^r}x^{13-3r}$$

For $$x^{-5}$$: $$13 - 3r = -5 \Rightarrow r = 6$$

Coefficient = $$\binom{13}{6}\frac{a^7}{b^6} = 1716\frac{a^7}{b^6}$$

Setting coefficients equal:

$$78\frac{a^{11}}{b^2} = 1716\frac{a^7}{b^6}$$

$$78 a^4 b^4 = 1716$$

$$a^4 b^4 = \frac{1716}{78} = 22$$

The correct answer is Option 3: 22.