Since $X$ takes values in $\{ 0, 1 \}$, then $e^{i t X}$ takes values in $\{ e^{i t \cdot 1}, e^{i t \cdot 0} \} = \{ e^{i t}, 1 \}$. Applying R1814: Expectation of discrete random euclidean real number, one then has \begin{equation} \begin{split} \mathbb{E}(e^{i t X}) & = e^{i t \cdot 1} \mathbb{P}(X = 1) + e^{i t \cdot 0} \mathbb{P}(X = 0) \\ & = e^{i t} \theta + 1 \cdot (1 - \theta) \\ & = \theta e^{i t} + 1 - \theta \\ \end{split} \end{equation} $\square$