von Bortkiewicz considered the frequency of deaths from kicks in the Prussian army corps. From the study of 14 corps over a 20-year period, he obtained the data shown in the table below. Fit a Poisson distribution to this data.
$$\begin{array}{|c|c|} \hline & \text{Number of deaths} & \text{Number of corps with x deaths in a given year} \\ \hline &0 & 144 \\ \hline &1 & 91\\ \hline &2 & 32 \\ \hline &3 & 11 \\ \hline &4 & 2 \\ \hline \end{array}$$ The solution computes $\lambda = 1\frac{91}{280}+2\frac{32}{280}+3\frac{11}{280}+4\frac{2}{280} = 0.7$ and the expected number of corps with x death as $280P(X = j)$ where P is a Poisson distribution with $\lambda = 0.7$ and X representing the number of death.
I see $\lambda$ as the expected number of death per corp-year (as it appears to be total death divided by total corps-year). If so, since $\lambda = np$, qualitatively what would $n$ and $p$ represent? I initially thought n is a corps-year and p is the probability of death, but this doesn't make sense as if each corps-year is a trial then X would be number of corps-year with a death, not number of death in a corps-year. To me $P(X =j)$ should give the probability of $j$ death in a corps-year.
You are correct that $\lambda$ is the Poisson rate of deaths per corps-years of exposure. So for a single corps${}^*$, the expected number of deaths in a single year is $0.7$.
Consequently, if $X$ is a random variable denoting the random number of deaths observed in a single corps in a year, then $X \sim \operatorname{Poisson}(\lambda)$ and $$\Pr[X = x] = e^{-\lambda} \frac{\lambda^x}{x!}, \quad x = 0, 1, 2, \ldots.$$ Now, among a sample of $n = 280$ corps, let $Y$ represent the number of corps with exactly $j$ deaths. More specifically, suppose we have $X_1, X_2, \ldots, X_{280} \sim X$ be independent and identically distributed as $X$, where $X_i$ is the random number of deaths in corps $i$ in one year, and $$Y_i = \mathbb{1} (X_i = j)$$ be an indicator random variable that equals $1$ if corps $i$ has exactly $j$ deaths that year. Then $$Y = Y_1 + Y_2 + \cdots + Y_{280}.$$ It is clear that $$\Pr[Y_i = 1] = \Pr[X_i = j] \overset{\text{iid}}{=} \Pr[X = j].$$ So $$Y \sim \operatorname{Binomial}(n = 280, p = \Pr[X = j])$$ and the expectation of $Y$ is $$\operatorname{E}[Y] = np = 280\Pr[X = j]$$ as claimed. Of course, it bears reminding that this is not to say that the expected number of deaths among the $280$ corps is equal to the expected number of corps with $j$ deaths.
${}^*$ The singular and plural of "corps" are both "corps."