We know that the Poisson distribution is given by

$p\left(x\right)=\frac{e^{-\lambda }{\lambda }^x}{x!}$

where, $x=0, 1, 2, 3, 4$

Since, $e^{-\lambda }$ is a positive constant therefore it is sufficient to find the maxima of $\left(\frac{{\lambda }^x}{x!}\right)$.

Now, when $x=0\Rightarrow \left(\frac{{\lambda }^x}{x!}\right)=1$

Now, we have to find maxima of $\left(\frac{{\lambda }^x}{x!}\right)$ for $x=1, 2, 3, 4, ....$

When $x=1\Rightarrow \left(\frac{{\lambda }^x}{x!}\right)=\left(\frac{{\lambda }^1}{1!}\right)=$

Now, we know that ${\log }_{e}y, y>0$ is an increasing function. Therefore, maximum of $\left(\frac{{\lambda }^x}{x!}\right)$ occurs when maximum of ${\log }_e\left(\frac{{\lambda }^x}{x!}\right)$ occurs. Now,

${\log }_e\left(\frac{{\lambda }^x}{x!}\right)=x{\log }_e\lambda -{\log }_e\left(x!\right)$

                    $=x{\log }_e\lambda -\left\{{\log }_e\left(x\right)+{\log }_e\left(x-1\right)+{\log }_e\left(x-2\right)+...+{\log }_{e}2+{\log }_{e}1\right\}$

                   $=x{\log }_e\lambda -\sum _{i=1}^x{\log }_e\left(i\right)$

           $=\left({\log }_e\lambda -{\log }_{e}1\right)+\left({\log }_e\lambda -{\log }_{e}2\right)+\left({\log }_e\lambda -{\log }_{e}3\right)+....+\left({\log }_e\lambda -{\log }_{e}x\right)$

Now, from above it is clear that when $x=1 ,2, 3$ then ${\log }_e\left(\frac{{\lambda }^x}{x!}\right)>0$ but if we start taking terms from $x=4, 5, 6, ....$ then ${\log }_e\left(\frac{{\lambda }^x}{x!}\right)<0$. Hence, the function start minimizing. So, maximum value of $\left(\frac{{\lambda }^x}{x!}\right)$ occurs when $x=3$.