$\def\a{\mathfrak{a}}
\def\b{\mathfrak{b}}
\def\c{\mathfrak{c}}
\def\q{\mathfrak{q}}
\def\g{\mathfrak{g}}
\def\p{\mathfrak{p}}
\def\ass{\operatorname{Ass}}$I will give a detailed argument to show equality of multiplicities of the associated ideals. The following text is a development of the ideas in CJD's answer.

From the hint (proven by Prof. Emerton) plus some argument (cf. [Y, Exercise 7.19]) one deduces

($*$) If $\bigcap_{i=1}^r \b_i=\bigcap_{j=1}^s \c_j$ are minimal irreducible decompositions, then $r=s$.

Let $\{\p_1,\dots,\p_n\}$ be the primes belonging to $\a$. We will denote this set as $\ass\a$ (although we won't ever need this, the notation means that the $\p_i$'s are the primes associated to the $A$-module $A/\a$; cf. [AM, Exercise 7.18]). We might relabel the $\b_i$'s and the $\c_j$'s as $\b_{ij}$, $\c_{ik}$, where $r(\b_{ij})=\p_i=r(\c_{ik})$, $1\leq j\leq r_i$, $1\leq k\leq s_i$. Then
$\q_i=\bigcap_{j=1}^{r_i}\b_{ij},\;
\g_i=\bigcap_{k=1}^{s_i}\c_{ik}$
are $\p_i$-primary ideals [AM, 4.3] and $\a=\bigcap_{i=1}^n\q_i=\bigcap_{i=1}^n\g_i$ are minimal primary decompositions. We want to show that $r_i=s_i$.
Let's proceed by induction in $\ell=$ maximal length of a chain in $\ass\a$. If $\ell=1$ that means every prime belonging to $\a$ is minimal. Thus every subset of $\ass\a$ is isolated (a subset of some poset is *isolated* if it is downward-closed) and $\q_i=\g_i$ by [AM, 4.10]. That is, we have an equality $\bigcap_{j=1}^{r_i}\b_{ij}=\bigcap_{k=1}^{s_i}\c_{ik}$ of minimal irreducible decompositions, whence $r_i=s_i$, by ($*$). Now suppose $\ell\geq 2$ and assume the result true for any ideal of $A$ whose maximal length of chains in the poset of its belonging primes is less than $\ell$. Let $T=\{\p_{i_1},\dots,\p_{i_m}\}$ be the primes sitting at the top of the chains of length $\ell$ in $\ass\a$. The sets $\Sigma=\ass\a\setminus T$ and $\Sigma_e=\Sigma\cup\{\p_{i_e}\}$ are isolated in $\ass\a$. Hence, by [AM, 4.10],
$$
\bigcap_{\p_i\in\Sigma}\q_i=\bigcap_{\p_i\in\Sigma}\g_i,\quad
\bigcap_{\p_i\in\Sigma_e}\q_i=\bigcap_{\p_i\in\Sigma_e}\g_i.
$$
From the induction hypothesis applied to the former ideal equality, we deduce $r_i=s_i$ for all $1\leq i\leq n$ such that $\p_i\not\in T$. From ($*$) applied to the latter ideal equality, we get $r_{i_e}+\sum_{\p_i\not\in T}r_i=s_{i_e}+\sum_{\p_i\not\in T}s_i$, whence $r_{i_e}=s_{i_e}$.

Lastly, I would like to note that the last sentence in the statement of [AM, Exercise 7.19] (omitted by OP) is “state and prove and analogous result for modules.” The generalization can be done for a Noetherian module (the base ring doesn't need to be Noetherian). This is written in [Y, Exercise 7.19], except that Yu's argument doesn't account the proof of $r_i=s_i$. The above argument is directly generalized to the module case, but instead now one invokes the corresponding generalized results coming from [AM, Exercises 4.22, 4.23] (you may read them in [Y]).

#### References

**AM.** Atiyah, Macdonald, *Introduction to Commutative algebra*.

**Y.** B. Yu, *Supplement and Solution Manual for Introduction
to Commutative algebra*. Also saved in the Wayback Machine.