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Problem is:

Let $R$ be a commutative ring and let $P$ be a prime ideal.

(a) Prove that the set of non-units in $R_P$ is the ideal $P_P$.

(b) Deduce that $R_P$ has a unique maximal ideal.

I tried (a) as:

Let $u$
be a non-unit in $R_p$
and $p$
be any elements in $R_p.$
Then, since $u$
is not unit, $upneq1$
and $puneq1.$
Suppose that $up$
or $pu$
is a unit in $R_p.$
Then there exists $q$
such that, say, $upq=1.$
Then $u(pq)=1.$
So, $u$
is a unit in $R_p.$
This is contradiction. So, the set of non-units in $R_p$
is an ideal in $R_P.$

So, I have to show that the set of non-units is equal to $P_P$ and (b)

But I am stuck at this point.

The ideal $P_P$ is
$$
P_P=leftfracas: ain P, sin Rsetminus Pright
$$
Let's prove that no element of $P_P$ is a unit. Suppose $ain P$, $sin S=Rsetminus P$, $xin R$ and $tin S$ be such that
$$
fracasfracxt=frac11
$$
By definition there exists $uin S$ with $u(ax-st)=0$ or $uax=ust$. This is a contradiction because $uaxin P$, but $ustin S$.

On the other hand, every element of $R_P$ not in $P_P$ is a unit. Indeed, if $a/snotin P_P$, we have $anotin P$, so $(a/s)^-1=s/a$.

Therefore the set of nonunits is an ideal and the ring is local, by general results. The maximal ideal is the set of nonunits, which we proved to be $P_P$.

Let $xin R$, denote by $[x]$ the class of $x$ in $R_P$, let $[M]=[p],pin P$,

$M$ is an ideal, if $[x],[y]in in M, [z]in R_P$ $[x]+[y]=[x+y]in M$, $[zx]in M$ since $x+y, zxin P$

$M$ is distinct of $R_P$, suppose $1in M$, this implies $1=pin P$ impossible since $P$ is an ideal.

$M$ is maximal. Let $N$ an ideal which contains strictly $M$, there exists $[s]in N$, $[s]$ is not in $M$ this implies $sin R-P$ thus $[s]$ is invertible and $1in N$.

Part (a) is Lemma 5.20 of the Sharp's book, "Steps in Commutative Algebra" (page 89):

It is easy to see that $I$ is an ideal of $R_P$.
Sharp proves that $R_psetminus I$ is the set of units. One direction is easy: for any $a/snotin I,$ one has $ain S$, so $dfracasdfracsa=dfrac11.$

On the other hand, if $b/t$ is a unit, say $dfracbtdfraccv=dfrac11,$ then there exists $win S$ such that $wbc=wtvin Rsetminus P.$ Hence $bnotin P,$ which means $b/tnotin I.$

For part (b), note that a ring is local if and only if the set of non-units of the ring is an ideal (this is lemma 3.13 of the book).