When the brick reaches 12 m height when thrown with velocity 12 m/s it has both potential energy and kinetic energy
When it just reaches the same height, it has only Potential energy, as its velocity is zero at its maximum height.
Therefore $ \%$ of energy saved $=\large\frac{KE}{KE+PE}$$\times 100$
$\qquad= \large\frac{\Large\frac{1}{2}mv^2}{\Large\frac{1}{2}mv^2+mgh} $$\times 100$
$\qquad=\large\frac{12^2}{12^2+2 \times 9.8 \times 12}$$ \times 100$
$\qquad=38 \%$
Hence b is the correct answer.