If $\quad X \sim B(n, p)$ then $\frac{P(X=k)}{P(X=k-1)}=$

For a binomial random variable $X \sim B(n, p)$, the probability mass function is $P(X=k) = \binom{n}{k} p^k q^{n-k}$ with $q = 1 - p$.

The ratio $\frac{P(X=k)}{P(X=k-1)}$ can be expressed as $\frac{\frac{n!}{k!(n-k)!} p^k q^{n-k}}{\frac{n!}{(k-1)!(n-k+1)!} p^{k-1} q^{n-k+1}}$.

Simplifying by canceling $n!$ and applying exponent rules, the expression reduces to $\frac{(k-1)!(n-k+1)!}{k!(n-k)!} \cdot \frac{p}{q}$.

Substituting $k! = k \cdot (k-1)!$ and $(n-k+1)! = (n-k+1) \cdot (n-k)!$ yields $\frac{n-k+1}{k} \cdot \frac{p}{q}$.

This corresponds to option D, confirming the final answer as $\boxed{\text{D}}$.