Answer
$P_2 = 0.91 $
Work Step by Step
Let $E_1 = 63 MeV + E_F$
and $E_2 = -63 MeV + E_F$
Use the probability formula
$ P_1 = \frac{1}{e^{E_1 - E_F/kT} + 1}$
$ P_1 = \frac{1}{e^x + 1}$
Solve for x,
$ (e^x + 1 )P_1 =1$
$e^xP_1 + P_1 =1$
$e^x=\frac{1-P_1}{P_1} $
$e^x = \frac{1}{P_1} - 1$
$e^x = \frac{1}{0.090} - 1$
$e^x = \frac{91}{9}$
Let's go back to the probability equation to find $P_2$
$ P_2 = \frac{1}{e^{E_2 - E_F/kT} + 1}$
But here, we use $E_2 - E_F = -63 MeV $
so $ -(E_2 - E_F)$
$ P_2 = \frac{1}{e^{-(E_2 - E_F)/kT} + 1}$
$ P_2 = \frac{1}{e^{-x} + 1}$
substitute $e^x = \frac{91}{9}$
$ P_2 = \frac{1}{(91/9){^-1} + 1}$
$ P_2 = 0.91 $
Addition : You can also validate this answer by adding both the probabilities to make sure it is approximately 1.00
$0.91 + 0.091 = 1.00$
