The foci of the ellipse \( \dfrac{x^{2}}{16} + \dfrac{y^{2}}{b^{2}} =1 \) and the hyperbola \( \dfrac{x^{2}}{144} - \dfrac{y^{2}}{81} =\dfrac{1}{25} \) coincide, then the value of \( b^{2} \) is:

A \( 5 \)

B \( 7 \)

C \( 9 \)