$\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$
$b^2=a^2(e^2-1)$
(or) $e^2=1+\large\frac{a^2+b^2}{a^2}$
$\qquad= \large\frac{a^2+b^2}{a^2}$
Conjugate hyperbola $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=-1$
(or) $\large\frac{y^2}{b^2}-\frac{x^2}{a^2}$$=1$
$\therefore e^{'2}=1+\large\frac{a^2}{b^2}$
$\qquad= \large\frac{a^2+b^2}{b^2}$
$\therefore \large\frac{1}{e^2} +\frac{1}{e^{'2}}=\frac{a^2+b^2}{a^2+b^2}$
$\qquad=1$
Hence B is the correct answer.