Let's find the eccentricities of the given ellipse and hyperbola, and then determine the eccentricity of an ellipse that passes through all four foci.

Step 1: Find $ e_1 $ for the Ellipse

The equation of the ellipse is:

$ \frac{x^2}{b^2} + \frac{y^2}{25} = 1 $

The eccentricity $ e_1 $ is given by:

$ e_1^2 = 1 - \frac{b^2}{25} $

Step 2: Find $ e_2 $ for the Hyperbola

The equation of the hyperbola is:

$ \frac{x^2}{16} - \frac{y^2}{b^2} = 1 $

The eccentricity $ e_2 $ is given by:

$ e_2^2 = 1 + \frac{b^2}{16} $

Step 3: Using the Product $ e_1 e_2 = 1 $

Given:

$ e_1 e_2 = 1 $

Thus:

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

Expanding gives:

$ 1 + \frac{b^2}{16} - \frac{b^2}{25} - \frac{b^4}{400} = 1 $

Simplifying:

$ \frac{9b^2}{400} = \frac{b^4}{400} $

Thus:

$ b^2 = 9 $

Step 4: Determine Eccentricities $ e_1 $ and $ e_2 $

Substitute $ b^2 = 9 $:

For the ellipse:

$ e_1^2 = 1 - \frac{9}{25} = \frac{16}{25} $

$ e_1 = \frac{4}{5} $

For the hyperbola:

$ e_2 = \frac{5}{4} $

Step 5: Find the Eccentricity of the New Ellipse

The new ellipse's equation is:

$ \frac{x^2}{25} + \frac{y^2}{16} = 1 $

The eccentricity $ e $ is:

$ e = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} $

Thus, the eccentricity of the ellipse that passes through all four foci is $ \frac{3}{5} $.