binomial-Poisson/beta hierarchy X|N,P ~ Binomial(N, P)
N ~ Poisson(11)
P ~ Beta(2,3)
What is the moment generating function for X?
 A: You've got the right start by conditioning on $N$ and $P$ to get $E[e^{tX}] = E[(Pe^t + 1-P)^N]$.  Now, evaluate the second expression via
$$E[(Pe^t + 1-P)^N] = E[E[(Pe^t + 1-P)^N|P]].$$
Since $N$ is Poisson, you will be able to get a closed-form answer for the inner expectation.  The only remaining variable will then be $P$.  But since $P$ is beta with small integer parameters the outer expectation will involve an integral of a low-degree polynomial times an exponential.  That can be evaluated via integration by parts.

(Update with more details, after comments below.  This problem is more complicated than I had realized at first!) 
Applying the calculations described above, we have, where $\lambda = 11$,
$$E[(Pe^t + 1-P)^N|P] = e^{\lambda P(e^t-1)}.$$
Then $E[e^{\lambda P(e^t-1)}]$ must be done piecewise, depending on the value of $t$.  If $t = 0$, we have
$$E[e^{\lambda P(e^t-1)}] = E[1] = 1.$$
Otherwise, $E[e^{\lambda P(e^t-1)}]$ is a much more complicated expression with, as the OP indicates, $(e^t-1)^4$ in the denominator:
$$E[e^{\lambda P(e^t-1)}] = \frac{12 \left(6 - 6 e^{\lambda (e^t-1)} + 2 \lambda (2 + e^{\lambda (e^t-1)}) (e^t-1) + 
   \lambda^2 (e^t-1)^2\right)}{\lambda^4 (e^t-1)^4}$$
Because the limit of this latter expression as $t \to 0$ is $1$, $E[e^{\lambda P(e^t-1)}]$ is still continuous at $0$, despite the fact that it has to be calculated piecewise.  It also turns out to be differentiable at $t = 0$ as well, and so one can still find moments with it.  For example, to find $E[X]$ we differentiate the expression for $E[e^{\lambda P(e^t-1)}]$, $t \neq 0$, once with respect to $t$ and then take the limit as $t \to 0$.  This gives $E[X] = 4.4$, which is the same result one would obtain by calculating $E[X]$ directly.
Incidentally, a similar situation occurs with the continuous uniform distribution on $[a,b]$.  Its mgf is, for $t \neq 0$, $$\frac{e^{tb}-e^{ta}}{t(b-a)}.$$  The moments can be found by successively differentiating this expression and then taking the limit as $t \to 0$ rather than by substituting $0$ for $t$.  See, for example, MathWorld's article on the uniform distribution. 
