Compute lambda of Poisson distribution $p = \lambda e^{ -\lambda }$
If $p$ and $e$ are known. How can I calculate $\lambda$?
I tried on log on both side but it did not help. Any suggestions?

You say that $e$ is known -- does that mean you're using $e$ as a variable name rather than to denote Euler's number? That would be rather confusing.
In either case, what you need is the Lambert W function. Using the defining relation
$$z=W(z)\mathrm e^{W(z)}$$
(where $\mathrm e$ as usual denotes Euler's number) and
$$-p=-\lambda\mathrm e^{-\lambda}\;,$$
you get
$$\lambda =-W(-p)\;.$$
Note that this is a) multivalued and b) only defined for $p\le1/e$.
If you really did intend to use $e$ as a variable name, we can replace it by $a$ to keep things clear and then write
$$
\begin{eqnarray}
p 
&=&
\lambda a^{-\lambda}
\;,
\\
p
&=&
\lambda \mathrm e^{-\lambda\log a}
\;,
\\
-p\log a
&=&
-\lambda\log a \mathrm e^{-\lambda\log a}
\;,
\\
\lambda
&=&
-\frac{W(-p\log a)}{\log a}
\;.
\end{eqnarray}$$
