Answer
$\color{blue}{\frac{2}{3}e^{-2} \approx 0.090}$
Work Step by Step
Let $X\sim \text{Poisson}(\lambda) \implies p_X(x) = \dfrac{e^{-\lambda}\lambda^x}{x!},\ x=0,1,2,3,\ldots;\ \lambda\gt 0$.
$P(X=1) = P(X=2) \\
\implies p_X(1) = p_X(2) \\
\implies \dfrac{e^{-\lambda}\lambda^1}{1!} = \dfrac{e^{-\lambda}\lambda^2}{2!} \\
\implies \lambda = \dfrac{\lambda^2}{2} \\
\implies \lambda^2-2\lambda = 0 \\
\implies \lambda(\lambda-2) = 0 \\
\implies \lambda=0, 2 \\
\implies \lambda = 2, \qquad \text{since }\lambda \gt 0.
$
Thus, if $X\sim \text{Poisson}(\lambda)$ and $P(X=1)=P(X=2)$, then $\lambda=2$.
Consequently, the pdf for such an $X$ is
$p_X(x) = \dfrac{e^{-2}2^x}{x!},\ x=0,1,2,3,\ldots$.
Thus,
$\begin{align*}
P(X=4) &= p_X(4) \\
&= \dfrac{e^{-2}2^4}{4!} \\
&= \dfrac{2}{3}e^{-2} \\
\color{blue}{P(X=4)} &\color{blue}{\approx 0.090}
\end{align*}$
