To find the difference between the eccentricities of the given hyperbola and the ellipse, we need to first understand their respective equations and calculate their eccentricities.

1. Consider the equation of the ellipse: \(px^2 + 16y^2 = 16p\)
   • Dividing the entire equation by \(16p\), we get: \(\frac{x^2}{\frac{16}{16}} + \frac{y^2}{\frac{p}{p}} = 1\), which simplifies to: \(\frac{x^2}{1} + \frac{y^2}{\frac{p}{16}} = 1\)
   • The standard form of an ellipse is: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)
   • From the simplified equation, \(a^2 = 1\) and \(b^2 = \frac{p}{16}\)
   • The eccentricity of the ellipse is given by: \(e_1 = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{p}{16}}\)
2. Consider the equation of the hyperbola: \(25(81x^2 - 144y^2) = 11664\)
   • Dividing the whole equation by \(11664\), we get: \(\frac{81x^2}{\frac{11664}{25}} - \frac{144y^2}{\frac{11664}{25}} = 1\), which simplifies to: \(\frac{x^2}{16} - \frac{y^2}{9} = 1\)
   • The standard form of a hyperbola is: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)
   • From the simplified equation, \(a^2 = 16\) and \(b^2 = 9\)
   • The eccentricity of the hyperbola is given by: \(e_2 = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}}\)
3. Calculate the eccentricities:
   • For the ellipse: \(e_1 = \sqrt{1 - \frac{p}{16}}\)
   • For the hyperbola: \(e_2 = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}\)
4. Given that the foci coincide, the values for \(e_1\) and \(e_2\) coincide with \(e_2\) calculated as \(\frac{5}{4}: (1.25)\). Hence, the noted difference between the eccentricities remains constant as:
   • Difference = \(e_2 - e_1\) = \(1.25\).

Thus, the difference between the eccentricities of the hyperbola and the ellipse is 1.25.