What is the NP? Suppose $X_1, X_2, X_3,\ldots, X_n$ are i.i.d. variables Poisson $(\lambda)$
and $g(λ)=\lambda(c - e^{-cλ})$
c:constant
What is the NP  for 
$H_0:g(λ)=c1$  vs $H_1:g(λ)=c2$      ??
My thought:
Step 1:  We prove  (monotone likelihood ratio property)
$$\forall \lambda_2>\lambda_1\:\: \frac{g(x\mid g(λ_2))}{g(x\mid g(λ_1))}$$
and (If I have not done any mistake) I proved
$\frac{g(x\mid g(λ_2))}{g(x\mid g(λ_1))}$ is non-decreasing (increasing) in 
$$T(X)=\sum(X_i)$$
Step 2 : after this point I am not sure how to continue the function hypothesis "$g(λ)$ confuses me .
 A: Let's start by observing that $g(\lambda)$ is a monotonically increasing function of $\lambda$.  This implies that for each value of $g(\lambda)$, there will be a unique value of $\lambda$; we can find a unique $\lambda$ such that $g(\lambda)=1$ and another such that $g(\lambda)=2$.  If we find these $\lambda$s, our problem reduces to finding the UMP test for the two point hypotheses associated with those two $\lambda$ values, and the Neyman-Pearson lemma assures us that such a test exists.
Unfortunately, there is no closed-form expression for the inverse function of $g(\lambda) = \lambda(1-e^{-\lambda})$, but we can find it numerically easily enough through any of a number of univariate root-finding algorithms, e.g., interval halving:
$$\lambda_1 = g^{-1}(1) \approx 1.35$$
$$\lambda_2 = g^{-1}(2) \approx 2.239$$
Defining $T(X) = \sum X_i$, as you have done, allows us to write the likelihood ratio for $H_1/H_0$ as:
$$\mathcal{L} = 2^{T(X)}e^{-1} \propto 2^{T(X)}$$
Taking the log of this last expression simplifies things, and is permissible as it is a monotonic transform of the likelihood ratio.  Our task then becomes finding the critical value $c$ for $T(X)$, where, under the null hypothesis, $T(X) \sim \text{Poisson}(g^{-1}(1)n) \approx \text{Poisson}(1.35n)$. If we do not care whether our test is exact, this amounts to finding $\min_c : P(c | n) \geq 1-\alpha$, e.g, if $n=10$ and $\alpha = 0.05$, then $c=20$, as $P(20|13.5)=0.965$ but $P(19|13.5) = 0.942$. 
Unfortunately, as $T(X)$ takes on only integer values, there may be no $c$ such that the probability that $T(X) > c$ under $H_0$ is exactly equal to our chosen test level $\alpha$.  If we wish an exact test, we will have to randomize.  
Following our example, this randomization results in the following criteria:
$\begin{align}
&\text{If}\,\, T(x) > 20, \text{reject} \,H_0 \\
&\text{If}\,\, T(x) < 20, \text{don't reject}\,H_0 \\
&\text{If}\,\, T(x) = 20, \text{reject}\, H_0\,\text{with probability}={0.95-0.942 \over 0.965-0.942} \approx 0.348
\end{align}$
where the randomization could occur by generating a $U(0,1)$ random number and comparing to the stated probability.
