Find unconditional distribution (using conditional expectation) $f(Θ)$ 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.
please answer..

I will distinguish between $\Theta$ and $\theta$ as described in the comments.  We have
\begin{align}
& \Pr(X=x) = \operatorname{E}(\Pr(X=x\mid\Theta)) = \operatorname{E}\left( \frac{\theta^x e^{-\theta}}{x!} \right) = \int_0^\infty \frac{\theta^x e^{-\theta}}{x!} f_\Theta(\theta)\,d\theta \\[10pt]
= {} & \frac 1 {x!} \int_0^\infty \theta^x e^{-\theta} \frac{\lambda^\alpha}{\Gamma(\alpha)}\theta^{\alpha-1}e^{-\lambda\theta} \, d\theta = \frac {\lambda^\alpha} {x!\Gamma(\alpha)} \int_0^\infty \theta^{x+\alpha-1} e^{-\theta(1+\lambda)} \, d\theta
\end{align}
The substitution $\eta = \theta(1+\lambda),\quad$ $\dfrac{d\eta}{1+\lambda} = d\theta$ then gets you some constant times a value of the Gamma function.
You will need to know why $\dfrac{\Gamma(x+\alpha)}{\Gamma(\alpha)} = \alpha(\alpha+1)(\alpha+2)\cdots(\alpha+x-1)$.
