**Rate Law Equation:**
The given rate equation is Rate = k[A][B], where [A] and [B] represent the concentrations of reactants A and B, and k is the rate constant.

**Effect of Volume Reduction:**
When the volume of the reaction vessel is reduced to 1/4th of its initial value, the concentration of the reactants will change accordingly. Let's assume the initial volume of the reaction vessel is V, and the new volume after reduction is V/4.

**Concentration Change:**
Since the total amount of substance remains constant, the concentration of the reactants will change inversely proportional to the change in volume. So, if the volume is reduced to 1/4th, the concentration will increase by a factor of 4.

For example, if the initial concentration of A is [A]0, then the new concentration of A after volume reduction will be 4[A]0.

**Rate Change:**
Substituting the new concentrations into the rate equation, the new rate (Rate') can be expressed as:

Rate' = k([A]')([B]') = k(4[A]0)(4[B]0) = 16k[A]0[B]0

Comparing the new rate with the initial rate (Rate), we can observe that the new rate is 16 times greater than the initial rate.

**Answer Explanation:**
According to the rate law equation and the effect of volume reduction, the new rate will be 16 times greater than the initial rate. Therefore, the correct answer is option D, which states that the new rate will be 16 (i.e., 16 times the initial rate).

Note: The answer options A, B, and C are incorrect because they do not reflect the correct relationship between the initial and new rates after the volume reduction.