B

Step 1: Calculate the initial kinetic energy of the bottom disc. The kinetic energy (K.E.) of a rotating object is given by the formula:
K.E. = \frac{1}{2} I \omega^2
where
I = moment of inertia of the disc = \frac{1}{2} m r^2
and \omega = angular velocity of the disc. For the bottom disc, the initial kinetic energy will be:
K.E._{initial} = \frac{1}{2} \left(\frac{1}{2} m r^2\right) \omega_0^2 = \frac{1}{4} m r^2 \omega_0^2.

Step 2: After the top disc sticks to the bottom disc, conservation of angular momentum must apply. The initial angular momentum of the system is:
L_{initial} = I_1 \omega_0 = \left(\frac{1}{2} m r^2\right) \omega_0
where I_1 is the moment of inertia of the bottom disc. The total moment of inertia of the two-disc system after the collision is:
I_{total} = I_1 + I_2 = \left(\frac{1}{2} m r^2 + \frac{1}{2} m r^2\right) = m r^2.

Step 3: The final angular velocity \omega_f can be found by setting the initial angular momentum equal to the final angular momentum:
L_{initial} = L_{final} => I_1 \omega_0 = I_{total} \omega_f
=> \left(\frac{1}{2} m r^2\right) \omega_0 = (m r^2) \omega_f.

Simplifying gives:
\omega_f = \frac{1}{4} \omega_0.

Step 4: Finally, calculate the final kinetic energy of the combined discs:
K.E._{final} = \frac{1}{2} \cdot I_{total} \cdot \omega_f^2 = \frac{1}{2} (m r^2) \left(\frac{1}{4} \omega_0\right)^2 = \frac{1}{2} (m r^2) \frac{1}{16} \omega_0^2 = \frac{m r^2 \omega_0^2}{32}.

Step 5: Calculate the ratio of kinetic energy before and after the collision:
\text{Ratio} = \frac{K.E._{initial}}{K.E._{final}} = \frac{\frac{1}{4} m r^2 \omega_0^2}{\frac{m r^2 \omega_0^2}{32}} = \frac{\frac{1}{4}}{\frac{1}{32}} = \frac{32}{4} = 8.

Thus, the final result is:
Therefore, the ratio of K.E. before and after the collision is 8:1. Hence the correct answer option is B.