My answer is just a detailed version of David Loeffler's comment.
First of all, I would observe that $N\cong \mathbb{Q}_p$ via
$$
\left(
\begin{array}{cc}
1&a\\
0&1
\end{array}\right)\mapsto a
$$
Moreover, since you look at compactly supported functions, each of them is in $\mathcal{C}^{sm}(p^{-k}\mathbb{Z_p},A)$ for some $k$, so $\mathcal{C}_c^{sm}(N,A)\cong\varinjlim_k\mathcal{C}^{sm}(p^{-k}\mathbb{Z}_p,A)$. Since injective limits commute with (almost everything and, in particular) homology, we are reduced to show
$$
\mathcal{C}^{sm}(\mathbb{Z}_p,A)_{p^k\mathbb{Z}_p}=0
$$
If we were considering simply the functions $\mathrm{Map}(\mathbb{Z}_p,A)$, the object in our hands would be $A[\mathbb{Z}_p]$. But you insist that your functions be smooth so in particular continuous which implies (and is hence equivalent) to being locally constant because $A$ is discrete. A locally constant function from $\mathbb{Z}_p$ to $A$ factors through $p^n\mathbb{Z}_p$ for some $n$ so we find
$$
\mathcal{C}^{sm}(\mathbb{Z}_p,A)=\varprojlim_nA[\mathbb{Z}/p^n\mathbb{Z}]\cong A[[T]]
$$
where the last isomorphism comes from sending a topological generator of the projective limit (like $(1,1,1,\dots)$) to $(1+T)$ (this is the Iwasawa algebra David was referring to). Keeping track of the action we see that it becomes multiplication by $(1+T)$, and the coinvariants with respect to $p^k\mathbb{Z}_p$ which we are looking for are the quotient $A[[T]]/(1+T)^k$ which vanishes because $(1+T)$ is a unit in the power series algebra.
