This is what I have so far for the problem:
There are 6 different ways to roll seven with 2 die, {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}. If you roll the two die n times, you have a $(\frac{1}{6})^n$ chance of rolling a seven. If we let $X_t$ be the process of rolling a die, $X_t ~$ Poisson$(\lambda)$. Let $X_t = k$ be the event you roll a seven.
Then, according to my professor, this is the rest of the problem. However, I don't understand how he got this distribution and simplified it.
Then, $P(X_t = k) = \sum_{n=0}^{\infty} (\frac{1}{6})^n * \frac{e^{\lambda t}(\lambda t)^n}{n!} = e^\frac{-5\lambda t}{6}$.
You have an error in your exponential. Its $e^{-\lambda t}$. Then you use conditional probabilities and denote the poisson process by $(N_t)_{t\geq 0}$:
$$P(X_t = k)=\sum_{n=0}^{\infty} P(X_t = k| N_t = n)P(N_t = n)=\sum_{n=0}^{\infty} \left(\dfrac{1}{6}\right)^n P(N_t = n)=\sum_{n=0}^{\infty} \left(\dfrac{1}{6}\right)^n \dfrac{e^{-\lambda t}(\lambda t)^n}{n!}$$
Now $ \left(\dfrac{1}{6}\right)^n (\lambda t )^n=\left(\dfrac{\lambda t}{6}\right)^n$. Hence,
$$\sum_{n=0}^{\infty} \left(\dfrac{1}{6}\right)^n \dfrac{e^{-\lambda t}(\lambda t)^n}{n!}=e^{-\lambda t}e^{\frac{\lambda t}{6}} = e^{-\frac{5\lambda t}{6}}.$$