Here is Steve D's argument (see comments). I hope Steve will post his own answer. This is a community wiki. [It's very possible of course that Steve's argument is correct but the way I "understood" it is not.]
I'll keep the notation $p$ but I'll only use the hypothesis $p > 1$ (and $p$ integer).
The group $C:=A/B$ is finitely generated. To prove that it is finite it suffices to show that its elements have finite order. Assume by contradiction there is a subgroup $D\subset C$ isomorphic to $\mathbb Z$. The multiplication by $p$ is surjective on $C$, since its image is exactly $p(A/B) = (pA + B)/B = A/B$. Let $E\subset C$ be the subgroup of all those $c$ in $C$ such that $p^nc$ is in $D$ for $n$ large enough. Then $E$ is a nonzero finitely generated torsion free abelian group on which the multiplication by $p$ is bijective. But no such group exists.
EDIT 1. Observation of Pete L. Clark (see Pete's comment below): Steve's argument shows that
(1) if $A$ is a finitely generated abelian group and $n > 1$ an integer satisfying $nA=A$, then $A$ is finite.
Of course this is obvious if one knows that a finitely generated abelian group is the direct sum of a free abelian group with a finite group.
My comment: One can also derive (1) from the result (2) below, stated as Corollary 2.5 in Atiyah-MacDonald, and as Theorem 47(a) in Pete's commutative algebra notes:
(2) Let $R$ be a ring, $J$ an ideal of $R$, and $M$ a finitely generated $R$-module such that $JM = M$. Then there exists $x\in R$ with $x\equiv 1\ (\bmod J)$ such that $xM = 0$.
As you can see in Pete's notes, (2) follows immediately from Cayley-Hamilton.
EDIT 2. Minor variation of the above: state the Nakayama Lemma as follows:
Let $R$ be a ring, $J$ an ideal, $M$ a finitely generated $R$-module and $N$ a submodule of $M$ such that $JM + N = M$. Then there exists $x\in R$ with $x\equiv 1\ (\bmod J)$ such that $xM\subset N$.
In particular, if $A$ is a finitely generated abelian group, $B$ a subgroup, $n > 1$ an integer satisfying $nA+B=A$, then there is a $k$ with $k\equiv 1\ (\bmod n)$ such that $kA\subset B$, showing that the index of $B$ is finite.
EDIT 3. For the sake of completeness, here is a proof of the structure theorem for finitely generated modules over a principal ideal domain, essentially taken from Algebraic Theory of Numbers by Pierre Samuel.
Let $A$ be a PID, $K$ its field of fractions, and $S$ a submodule of $A^n$. The maximum number of linearly independent elements of $S$ is also the dimension of the vector subspace of $K^n$ generated by $S$. Thus this integer, called the rank of $S$, only depends on the isomorphism class of $S$ and is additive with respect to finite direct sums.
Theorem. (a) $S$ is free of rank $r\le n$.
(b) There is a basis $u_1,\dots,u_n$ of $A^n$ and there are nonzero elements $a_1,\dots,a_r$ of $A$ such that $a_1u_1,\dots,a_ru_r$ is a basis of $S$ and $a_1 | a_2 | \cdots | a_r$, where $a | b$ means “$a$ divides $b$”.
We'll need the
Proposition. Let $f$ be an $A$-valued bilinear map defined on a product of two $A$-modules. Then the image of $f$ is an ideal.
Proof of the Proposition. Let $T$ be the set of all ideals of the form $A f(x,y)$, let $A f(x,y)$ and $A f(u,v)$ be two elements of $T$, and suppose that $A f(x,y)$ is maximal in $T$. It suffices to show that $f(x,y) | f(u,v)$.
Claim: $f(x,y) | f(x,v)$ and $f(x,y) | f(u,y)$. Indeed, any generator of $Af(x,y)+Af(x,v)$ is of the form $$a f(x,y)+bf(x,v)=f(x,ay+bv),$$ and the maximality of $A f(x,y)$ implies $f(x,y) | f(x,v)$. The proof of $f(x,y) | f(u,y)$ is similar.
We can assume $f(x,v)=0=f(u,y)$. [Write $f(x,v)=a f(x,y)$ and replace $v$ by $v-ay$. Similarly for $u$.] Using the equality $$f(a x+b u,y+v)=a f(x,y)+b f(u,v)$$ for all $a$ and $b$ in $A$, and arguing as in the proof of the claim, we see that $f(x,y) | f(u,v)$. QED
Proof of the Theorem. We assume (as we may) that $S$ is nonzero, we let $f$ be the bilinear form on $A^n$ whose matrix with respect to the canonical basis is the identity matrix, and we pick a generator $a_1=f(s_1,y_1)$ of $f(S\times A^n)$. [Naively: $a_1$ is the gcd of the coordinates of the elements of $S$.] Clearly, $u_1:=s_1/a_1$ is in $A^n$, and we have
$$A^n=Au_1\oplus y_1^\perp,\quad S=As_1\oplus(S\cap y_1^\perp),$$
where $y_1^\perp$ is the orthogonal of $y_1$. Then (a) follows by induction on $r$. In particular $y_1^\perp$ and $S\cap y_1^\perp$ are free of rank $n-1$ and $r-1$, and (b) follows also by induction. [Note that if $x$ belongs to a basis of $A^n$, then $f(x,y)=1$ for some $y$ in $A^n$.] QED
I don’t know if there is a simple characterization of the rings for which the Proposition holds. (See the Wikipedia entry for Principal ideal rings.)
PDF version of this edit
EDIT 4. A nice application of Nakayama's Lemma (see Theorem 52 p. 48 in Pete L. Clark's commutative algebra notes):
Let $B$ be a commutative ring and $V$ a finitely generated $B$-module.
We start with the following form of Nakayama's Lemma:
If $J$ is an ideal of $B$ satisfying $JV=V$, then $xV=0$ for some $x$ in $B$ with $x\equiv 1$ mod $J$.
Corollary:
If in addition $v$ is in $V$ and satisfies $Jv=0$, then $v=0$.
Subcorollary:
The action of $b\in B$ on $V$ is injective if surjective.
To see this, put $J:=Bb$.
Subsubcorollary:
Let $A$ be a commutative ring, $V$ a finitely generated $A$-module, and $b$ an endomorphism of $V$. Then $b$ is injective if surjective.
To see this, put $B:=A[b]$.
This gives a solution to Exercise 3.15 in Introduction to Commutative Algebra by Atiyah and MacDonald.
Best Answer
Since $N$ is finitely generated and of finite index, $G$ itself is finitely generated (by generators of $N$, moved by the finitely many coset representatives), so $G\cong \mathbb{Z}^r\oplus \Delta$, for a finite group $\Delta$. Since $G/N$ is finite, $N$ has maximal rank $r$, which answers your question affirmatively.
But be careful: it is not true that there is necessarily a subgroup $F$ such that $G=N\oplus F$. A simple counterexample is given by $2\mathbb{Z}\subset \mathbb{Z}$. The quotient is cyclic of order $2$ but the subgroup if not complemented, i.e. the short exact sequence does not split. Of course, $\mathbb{Z}$ is still isomorphic to $2\mathbb{Z}\oplus\{1\}$.