Till $\theta < \tan ^{-1} \mu$ contact force is constant and gravitational force are equal and opposite.
Once the body starts sliding
$F_c=\sqrt {(mg \cos \theta)^2+(\mu \;mg\; \cos \theta)^2}$
$\quad= mg \cos \theta \sqrt {1+\mu ^2}$
So as $\theta $ increases $\cos \theta$ decreases.
So contact force decreases.
Hence b is the correct answer.