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$f(Θ)$ is pdf of gamma distribution
$$f(Θ) = \frac{λ^α}{Γ(α)}Θ^{α-1}\exp(-λΘ), $$
$$X\mid Θ \sim \mathrm{poisson}(Θ) \rightarrow \frac{Θ^x\exp(-Θ)}{x!}$$
Suppose that $Θ$ is a random variable that follows a gamma distribution with parameters $λ$ and $α$, where $α$ is an integer, and suppose that, conditional on $Θ$, $X$ follows a Poisson distribution with parameter $Θ$. Find the unconditional distribution of $α + X$ (Hint : Find the mgf by using iterated conditional expectations.)
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I got
$$M_{a+X} = \frac{(λexp(t))^α}{(λ-exp(t)+1)^α}$$
But I don't know distribution has this mgf.
What is the distribution ?
It seems easier to me, at least, to find the unconditional distribution directly: $$\begin{align*} \Pr[X = x] &= \int_{\theta = 0}^\infty \Pr[X = x \mid \Theta = \theta] f_\Theta(\theta) \, d\theta \\ &= \int_{\theta = 0}^\infty e^{-\theta} \frac{\theta^x}{x!} \frac{\lambda^\alpha \theta^{\alpha-1} e^{-\lambda \theta}}{\Gamma(\alpha)} \, d\theta \\ &= \int_{\theta = 0}^\infty \frac{\lambda^\alpha \theta^{x+\alpha-1} e^{-(\lambda+1)\theta}}{x! (\alpha-1)!} \, d\theta \\ &= \frac{(x+\alpha-1)!}{x! (\alpha-1)!} \frac{\lambda^\alpha}{(\lambda+1)^{x+\alpha}}\int_{\theta=0}^\infty \frac{(\lambda+1)^{x+\alpha} \theta^{x+\alpha-1} e^{-(\lambda+1)\theta}}{\Gamma(x+\alpha)} \, d\theta \end{align*}$$ and then recognizing that the integrand is the density of a gamma distribution with rate $\lambda + 1$ and shape $x+\alpha$, thus equals $1$. It follows that $$\Pr[X = x] = \binom{x+\alpha-1}{\alpha-1} \left(\frac{1}{\lambda+1}\right)^x \left(\frac{\lambda}{\lambda+1}\right)^\alpha$$ which is quite clearly negative binomial. I leave it to you to determine what the relevant parameters are; then $\alpha + X$ is also negative binomial, albeit with a different support and parametrization.
For the sake of completeness, we will do the calculation of the moment generating function. Recall that the MGFs of the Poisson and gamma distributions are, respectively $$M_{X \mid \Theta}(t) = \operatorname{E}[e^{tX}\mid\Theta] = \exp((e^t-1)\Theta),$$ and $$M_\Theta(m) = \operatorname{E}[e^{m\Theta}] = (1 - m/\lambda)^{-\alpha}$$ for the chosen shape and rate parametrization you wrote in your question. Consequently, $$M_X(t) = \operatorname{E}[e^{tX}] = \operatorname{E}[\operatorname{E}[e^{tX}\mid\Theta]] = \operatorname{E}[M_{X\mid\Theta}(t)] = \operatorname{E}[\exp((e^t-1)\Theta)] = M_\Theta(e^t-1) = \left(1 - \frac{e^t-1}{\lambda}\right)^{-\alpha}.$$ This of course is equivalent to $$M_X(t) = \left(\frac{\lambda/(1+\lambda)}{1 - e^t(1 - \lambda/(1+\lambda))}\right)^\alpha,$$ and now we can see that $p = \lambda/(1+\lambda)$ will give us the form of a negative binomial MGF.