1. Interpret the reaction

   Since only substances $A$ and $B$ are mentioned, the intended chemical reaction is the simple reversible reaction: $A \rightleftharpoons B$

   For this reaction, $K = \frac{[B]_{eq}}{[A]_{eq}}$

2. Let the initial concentration of $A$ be $a$

   Then the initial concentration of $B$ is given as $1.5$ times that of $A$: $[A]_0 = a, \qquad [B]_0 = 1.5a = \frac{3a}{2}$

3. Assume the reaction shifts backward

   Since initially $B$ is greater than $A$, and at equilibrium they become equal, some $B$ must convert to $A$.

   Let $x$ be the amount of $B$ converted to $A$.

   Then at equilibrium: $[A]_{eq} = a + x$ $[B]_{eq} = \frac{3a}{2} - x$

4. Use the condition that equilibrium concentrations are equal

   Given: $[A]_{eq} = [B]_{eq}$

   Therefore, $a + x = \frac{3a}{2} - x$

   $2x = \frac{a}{2}$

   $x = \frac{a}{4}$

5. Find the equilibrium concentrations

   $[A]_{eq} = a + \frac{a}{4} = \frac{5a}{4}$ $[B]_{eq} = \frac{3a}{2} - \frac{a}{4} = \frac{6a- a}{4} = \frac{5a}{4}$

   As expected, both are equal.

6. Calculate the equilibrium constant

   For $A \rightleftharpoons B$, $K = \frac{[B]_{eq}}{[A]_{eq}} = \frac{\frac{5a}{4}}{\frac{5a}{4}} = 1$

7. Check options

   • A: $16$ ❌
   • B: $1$ ✅
   • C: $\frac{1}{4}$ ❌
   • D: $4$ ❌

Hence, the correct answer should be: $\boxed{1}$