The question describes a fundamental limit theorem in probability theory.

We are given a sequence of random variables $X_n$ such that $X_n \sim Bin(n, p_n)$. This means $X_n$ follows a binomial distribution with $n$ trials and probability of success $p_n$ for each trial.
The conditions for convergence are $p_n \to 0$ as $n \to \infty$ and $np_n \to \lambda$ for some constant $\lambda > 0$.

The Poisson limit theorem states that if $X_n$ follows a binomial distribution $Bin(n, p_n)$, and as $n \to \infty$, $p_n \to 0$ such that the product $np_n$ approaches a finite positive constant $\lambda$, then $X_n$ converges in distribution to a Poisson distribution with parameter $\lambda$.

• A. Bernoulli ($\lambda$): A Bernoulli distribution is for a single trial, not a limit of $n$ trials.
• B. Degenerate at $\lambda$: A degenerate distribution means the random variable takes a single value with probability 1. This is not the case for the limit of a binomial distribution under these conditions.
• C. Normal ($\lambda, \lambda$): The normal approximation to the binomial distribution is typically used when $n$ is large and $p$ is not too close to 0 or 1. Here, $p_n \to 0$, which violates the conditions for a good normal approximation.
• D. Poisson ($\lambda$): This matches the statement of the Poisson limit theorem.

Therefore, $X_n$ converges in distribution to a Poisson distribution with parameter $\lambda$.

The final answer is $\boxed{\text{D}}$