Let $$e_1$$ and $$e_2$$ be the eccentricities of the ellipse $$\frac{x^2}{25} + \frac{y^2}{b^2} = 1$$ $$(b < 5)$$ and the hyperbola $$\frac{x^2}{16} - \frac{y^2}{b^2} = 1$$ respectively satisfying $$e_1 e_2 = 1$$. If $$\alpha$$ and $$\beta$$ are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair $$(\alpha, \beta)$$ is equal to:

We have an ellipse whose equation is $$\dfrac{x^2}{25} + \dfrac{y^2}{b^2} = 1,\; (b < 5).$$

For any ellipse of the form $$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$$ with its major axis along the $$x$$-axis, the standard results are

$$a^2 = \text{denominator of }x^2,\quad b^2 = \text{denominator of }y^2,$$

$$c^2 = a^2 - b^2,\quad e = \dfrac{c}{a},$$

where $$c$$ is the semi-focal distance and $$e$$ is the eccentricity.

Comparing, we identify $$a^2 = 25$$ and $$b^2 = b^2$$ itself. Therefore

$$c_1^2 = a^2 - b^2 = 25 - b^2,$$

so $$c_1 = \sqrt{25 - b^2}.$$

Hence the eccentricity of the ellipse is

$$e_1 = \dfrac{c_1}{a} = \dfrac{\sqrt{25 - b^2}}{5}.$$

Next we have a hyperbola whose equation is $$\dfrac{x^2}{16} - \dfrac{y^2}{b^2} = 1.$$

For any hyperbola of the form $$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1,$$ the corresponding formulas are

$$a^2 = \text{denominator of }x^2,\quad b^2 = \text{denominator of }y^2,$$

$$c^2 = a^2 + b^2,\quad e = \dfrac{c}{a},$$

because in a hyperbola the relationship involves a plus sign in $$c^2 = a^2 + b^2.$$

Here $$a^2 = 16$$ and again $$b^2 = b^2.$$ Thus

$$c_2^2 = a^2 + b^2 = 16 + b^2,$$

so $$c_2 = \sqrt{16 + b^2}.$$

The eccentricity of the hyperbola is therefore

$$e_2 = \dfrac{c_2}{a} = \dfrac{\sqrt{16 + b^2}}{4}.$$

The condition given in the problem is $$e_1 e_2 = 1.$$ Substituting the expressions we have just derived,

$$\left(\dfrac{\sqrt{25 - b^2}}{5}\right) \left(\dfrac{\sqrt{16 + b^2}}{4}\right) = 1.$$

Multiplying the numerators and denominators separately we get

$$\dfrac{\sqrt{(25 - b^2)(16 + b^2)}}{20} = 1.$$

Now multiply both sides by $$20$$:

$$\sqrt{(25 - b^2)(16 + b^2)} = 20.$$

To remove the square root we square both sides:

$$(25 - b^2)(16 + b^2) = 400.$$

Expanding the left-hand side step by step,

$$25 \times 16 + 25 b^2 - 16 b^2 - b^4 = 400.$$

The product $$25 \times 16 = 400,$$ so we have

$$400 + 25 b^2 - 16 b^2 - b^4 = 400.$$

Combining the like terms in $$b^2$$ gives

$$400 + 9 b^2 - b^4 = 400.$$

Subtract $$400$$ from both sides:

$$9 b^2 - b^4 = 0.$$

Factor out $$b^2$$:

$$b^2 (9 - b^2) = 0.$$

The factor $$b^2 = 0$$ is inadmissible because the denominators in the original equations would vanish. Therefore,

$$9 - b^2 = 0 \quad \Longrightarrow \quad b^2 = 9.$$

Since $$b > 0$$ and the problem states $$b < 5,$$ we choose

$$b = 3.$$

Now we can compute the quantities asked for. First, the distance between the two foci of the ellipse is

$$\alpha = 2 c_1 = 2 \sqrt{25 - b^2} = 2 \sqrt{25 - 9} = 2 \times 4 = 8.$$

Next, the distance between the two foci of the hyperbola is

$$\beta = 2 c_2 = 2 \sqrt{16 + b^2} = 2 \sqrt{16 + 9} = 2 \times 5 = 10.$$

Hence the ordered pair is $$(\alpha, \beta) = (8, 10).$$

Looking at the options, this matches Option A.

Hence, the correct answer is Option A.