Answer
$\color{blue}{\frac{1}{2}\left(1+e^{-2\lambda}\right)}$
Work Step by Step
$\underline{\text{A useful identity}}$
Note that for all real $x$,
$e^x = \displaystyle \sum_{k=0}^\infty \dfrac{x^k}{k!} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$
and
$e^{-x} = \displaystyle \sum\limits_{k=0}^\infty \dfrac{(-1)^kx^k}{k!} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$.
Adding these two power series cancels out the odd-powered terms and doubles the even-powered terms eventually giving:
$\begin{align*}
e^x +e^{-x} &= 2(1) + 2\frac{x^2}{2!} + 2\frac{x^4}{4!} + \cdots \\
&= \sum\limits_{k=0}^\infty 2\dfrac{x^{2k}}{(2k)!},
\end{align*}$
so that
$\dfrac{e^x+e^{-x}}{2} = \displaystyle \sum\limits_{k=0}^\infty \dfrac{x^{2k}}{(2k)!}. \qquad\qquad \text{(Eq. 1)}$.
$\underline{P(X\ \text{is even})}$
Now, suppose $X \sim \text{Poisson}(\lambda)$ so that the pdf of $X$ is
$p_X(k) = e^{-\lambda}\lambda^k/k!,\ k=0,1,2,3,\ldots$.
Then, the probability that $X$ is even is given by
$\begin{align*}
P( X\ \text{is even}) &= P(X=0) + P(X=2) + P(X=4) + \cdots \\
&= p_X(0) + P_X(2) + p_X(4) + \cdots \\
&= \frac{e^{-\lambda}\lambda^0}{0!} + \frac{e^{-\lambda}\lambda^2}{2!} + \frac{e^{-\lambda}\lambda^4}{4!} + \cdots \\
&= \sum_{k=0}^\infty \frac{e^{-\lambda}\lambda^{2k}}{(2k)!} \\
&= e^{-\lambda}\underbrace{\sum_{k=1}^\infty \frac{\lambda^{2k}}{(2k)!}}_{\text{in Eq. 1, let}\ x\ =\ \lambda} \\
&= e^{-\lambda}\left(\dfrac{e^{\lambda} + e^{-\lambda}}{2}\right) \\
&= \dfrac{e^{-\lambda}e^\lambda + e^{-\lambda}e^{-\lambda}}{2} \\
&= \dfrac{1+e^{-2\lambda}}{2} \\
\color{blue}{P( X\ \text{is even})} &\color{blue}{= \frac{1}{2}\left(1+e^{-2\lambda}\right)}
\end{align*}$
