The exercise is copied/paraphrased below.
Let $A$ be a ring with the following property.
(L1) For every ideal $\mathfrak{a}\neq (1)$ in $A$ and every prime ideal $\mathfrak{p}$, there exists $x\not\in\mathfrak{p}$ such that $S_{\mathfrak{p}}(\mathfrak{a})=(\mathfrak{a}:x)$. Here $S_{\mathfrak{p}}(\mathfrak{a})\subseteq A$ denotes the contraction of $\mathfrak{a}A_{\mathfrak{p}}$ along $A \rightarrow A_{\mathfrak{p}}$.
Then every ideal in $A$ is an intersection of (possibly infinitely many) primary ideals.
There is a detailed hint given. Fixing $\mathfrak{a}$, the hint lets $\mathfrak{p}$ be minimal among primes containing $\mathfrak{a}$, and constructs a $\mathfrak{p}$-primary ideal $\mathfrak{q}\supseteq \mathfrak{a}$ along with $x\not \in \mathfrak{p}$ such that $\mathfrak{a}=\mathfrak{q}\cap(\mathfrak{a}+(x))$.
One can iterate this construction with $\mathfrak{a}+(x)\supsetneq \mathfrak{a}$ replacing $\mathfrak{a}$. If this never terminates, we get (with a little more work) a strict ascending chain of ideals which can be indexed by a set with cardinality as large as we please; but this cannot be bigger than the cardinality of the ring $A$. As soon as it terminates, we have a (possibly infinite) primary decomposition.
Question: Rather than iterating with $\mathfrak{a}+(x)$, the book chooses to instead iterate with a maximal element of the set of ideals $\mathfrak{b}\supseteq \mathfrak{a}$ such that $\mathfrak{a}=\mathfrak{q}\cap\mathfrak{b}$. The ideal $\mathfrak{a}+(x)$ satisfies this condition but may not be maximal. Is the maximality assumption relevant to the solution?
It seems to me that maximality is not relevant, and one can just take $\mathfrak{a}+(x)$. Have I missed something?