Let $$e_1$$ and $$e_2$$ be the eccentricities of the ellipse $$\frac{x^2}{b^2} + \frac{y^2}{25} = 1$$ and the hyperbola $$\frac{x^2}{16} - \frac{y^2}{b^2} = 1$$, respectively. If $$b < 5$$ and $$e_1 e_2 = 1$$, then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is :

For the first ellipse $$\dfrac{x^{2}}{b^{2}}+\dfrac{y^{2}}{25}=1$$ the larger denominator is $$25$$, so the semi-major axis length is $$a_1 = 5$$ and the semi-minor axis length is $$b_1 = b$$.

Eccentricity of an ellipse is given by $$e=\sqrt{1-\dfrac{b^{2}}{a^{2}}}$$.
Hence $$e_1 = \sqrt{1-\dfrac{b^{2}}{25}} \quad -(1)$$

For the hyperbola $$\dfrac{x^{2}}{16}-\dfrac{y^{2}}{b^{2}}=1$$ the semi-transverse axis is $$a_2 = 4$$ and the semi-conjugate axis is $$b_2 = b$$.

Eccentricity of a hyperbola is given by $$e=\sqrt{1+\dfrac{b^{2}}{a^{2}}}$$.
Hence $$e_2 = \sqrt{1+\dfrac{b^{2}}{16}} \quad -(2)$$

Given $$e_1\,e_2 = 1$$. Substituting from $$(1)$$ and $$(2)$$:

$$\sqrt{1-\dfrac{b^{2}}{25}}\;\sqrt{1+\dfrac{b^{2}}{16}} = 1$$

Squaring both sides:

$$\left(1-\dfrac{b^{2}}{25}\right)\!\left(1+\dfrac{b^{2}}{16}\right)=1$$

Expanding and simplifying:

$$1+\dfrac{b^{2}}{16}-\dfrac{b^{2}}{25}-\dfrac{b^{4}}{400}=1$$
$$\dfrac{b^{2}}{16}-\dfrac{b^{2}}{25}-\dfrac{b^{4}}{400}=0$$
$$b^{2}\left(\dfrac{1}{16}-\dfrac{1}{25}\right)-\dfrac{b^{4}}{400}=0$$
$$b^{2}\left(\dfrac{9}{400}\right)-\dfrac{b^{4}}{400}=0$$
$$\dfrac{b^{2}}{400}\Bigl(9-b^{2}\Bigr)=0$$

Since $$b\neq 0$$, we get $$b^{2}=9 \Rightarrow b=3 \;(\text{given } b\lt 5).$$

Now compute the four focal points:

Eccentricity $$e_1 = \sqrt{1-\dfrac{9}{25}}=\dfrac{4}{5}$$.
Distance of each focus from the centre: $$c_1=a_1e_1=5\left(\dfrac{4}{5}\right)=4$$.
Ellipse foci: $$(0,\pm4).$$

Eccentricity $$e_2 = \sqrt{1+\dfrac{9}{16}}=\dfrac{5}{4}$$.
Distance of each focus from the centre: $$c_2=a_2e_2=4\left(\dfrac{5}{4}\right)=5$$.
Hyperbola foci: $$(\pm5,0).$$

The required ellipse has its axes along the coordinate axes and must pass through all four foci $$(\pm5,0),\,(0,\pm4)$$. Take its equation as

$$\dfrac{x^{2}}{A^{2}}+\dfrac{y^{2}}{B^{2}}=1.$$

Substituting point $$(5,0):\; \dfrac{25}{A^{2}} = 1 \Rightarrow A^{2}=25 \Rightarrow A=5.$$

Substituting point $$(0,4):\; \dfrac{16}{B^{2}} = 1 \Rightarrow B^{2}=16 \Rightarrow B=4.$$

Thus the ellipse is $$\dfrac{x^{2}}{25}+\dfrac{y^{2}}{16}=1$$ with semi-major axis $$a=5$$ and semi-minor axis $$b=4$$ (since $$25\gt16$$).

Eccentricity of this ellipse is

$$e = \sqrt{1-\dfrac{b^{2}}{a^{2}}} = \sqrt{1-\dfrac{16}{25}} = \sqrt{\dfrac{9}{25}} = \dfrac{3}{5}.$$

Therefore, the required eccentricity is $$\dfrac{3}{5}$$.

Option B is correct.