The prior "exponential with mean one" has density $\pi(\lambda)=e^{-\lambda}$ , $\lambda>0$ — that is $\text{Gamma}(\text{shape }1,\text{ rate }1)$ .

The likelihood from $n=4$ independent Poisson observations is

$L(\lambda)=\prod_{i=1}^{4}\frac{e^{-\lambda}\lambda^{k_i}}{k_i!} \propto e^{-4\lambda}\,\lambda^{\sum k_i}.$

Multiply by the prior and keep only $\lambda$ -terms:

$\pi(\lambda\mid\uline{k}) \propto\ e^{-4\lambda}\lambda^{\sum k_i}\cdot e^{-\lambda} =\lambda^{\sum k_i}\,e^{-5\lambda}.$

Comparing with the $\text{Gamma}(\alpha,\beta)$ kernel $\lambda^{\alpha-1}e^{-\beta\lambda}$ gives

$\alpha=\sum_{i=1}^{4}k_i+1, \qquad \beta = \underbrace{4}_{\text{from the data}}+\underbrace{1}_{\text{from the prior}}=5 .$

So the posterior is $\text{Gamma}\!\left(\sum k_i+1, 5\right)$ — the Gamma family is conjugate for the Poisson, the shape absorbing the total count and the rate absorbing the sample size.

Note — no option is exactly right. The correct rate is $5$ ; the choices offer $8$ and $4$ . Only (A) and (C) carry the correct shape $\sum k_i+1$ , and of these (C) has the rate $4$ , which is what you get by counting the four observations and forgetting the prior's own rate of $1$ . (C) is therefore taken as the intended key, but the fully correct posterior is $\text{Gamma}\!\left(\sum k_i+1, 5\right)$ .