Okay, let's look at Example 1.3 from the text: consider the affine curve $E: y^2 = x^3 + x$. This curve is smooth: letting $f(x,y) = y^2 - (x^3 + x)$, one can show there is no point where $f$ and both partial derivatives $f_x$ and $f_y$ vanish simultaneously.
Recall that the local ring $\overline{K}[E]_P$ can be defined as all rational functions on $E$ that are defined at the point $P$, i.e., with denominators that don't vanish at $P$. This ring is local hence has a unique maximal ideal $\mathfrak{m}_P$, which consists of all rational functions on the curve that vanish at $P$.
Since the curve is smooth, then the local ring $\overline{K}[E]_P$ is a discrete valuation ring for each point $P \in E(\overline{K})$. This means that the unique maximal ideal $\mathfrak{m}_P$ of $\overline{K}[E]_P$ is generated by a single element $t$, called a uniformizer. Usually, we pick $t$ to be some line that is not tangent to the curve at $P$: we don't want the tangent line because it "vanishes to order 2" at $P$. Once we've found a uniformizer, our lives are easier: in order to determine $\text{ord}_P(f)$ of a function $f$, we must find the highest power $d$ such that $f \in \mathfrak{m}_P^d$. But this is just the number of factors of $t$ in $f$: if $\text{ord}_P(f) = d$, then we can write $f = u t^d$ for some unit $u \in \overline{K}[E]_P$.
Let's consider the point $P = (0,0)$ on $E$, so $\mathfrak{m}_P = (x,y)$. We want to find a uniformizer for $\mathfrak{m}_P$. At the point $P = (0,0)$, the tangent line to $E$ is the line $x = 0$: one can see this because it is the lowest order term appearing in the "Taylor expansion" $0 = x^3 - y^2 + x$, or because $x = y^2 - x^3 \in \mathfrak{m}_P^2 = (x^2, xy, y^2)$. So instead, we can take $y$ as our uniformizer and $\mathfrak{m}_P = (x,y) = (y)$. Since $y$ has only one factor of $y$, then $\text{ord}_P(y) = 1$. Here are two ways to compute $\text{ord}_P(x)$:
1) Since $y^2 = x^3 + x = (x^2 + 1)x$, then we can write $x = \frac{y^2}{1+x^2}$. Since $1 + x^2$ doesn't vanish at $(0,0)$, then it is a unit, hence has order $0$ at $P$. Then we have $x = u y^2$, where $u = \frac{1}{1+x^2}$, so $\text{ord}_P(x) = 2$.
2) We have $x = y^2 - x^3$, so
$$
\text{ord}_P(x) = \text{ord}_P(y^2 - x^3) = \min\{\text{ord}_P(y^2), \text{ord}_P(x^3)\} = 2
$$
by the second property of the discrete valuation, since $\text{ord}_P(x^3) \geq 3$.
Now that you've seen these two examples, see if you can show $\text{ord}_P(2y^2 - x) = 2$ as stated in the text.
Given $g \in M_p$, then $g(p) = 0$ so $\DeclareMathOperator{\ord}{ord} \ord_p(g) \geq 1$. Since $\ord_p(t) = 1$, then
$$
\ord_p(g/t) = \ord_P(g) - \ord_p(t) = \ord_P(g) - 1 \geq 0
$$
so $g/t \in \overline{K}[C]_p$ since $\overline{K}[C]_p$ is the valuation ring of $\ord_p$. Then there exists a function $h \in \overline{K}[C]_p$ such that $g = ht$, so $g \in (t)$. Thus $M_p \subseteq (t)$ and the other inclusion is immediate.
(You certainly could use Nakayama's lemma to prove this. Since $M_p/M_p^2$ is $1$-dimensional and $t \notin M_p^2$, then $\{t\}$ is a basis for $M_p/M_p^2$.)
Best Answer
I think you're confusing $p$ with $P$. Assuming you meant $P$, your question follows from standard commutative algebra. As stated in the other answer, $\bar{K}[C]_P$ is the local ring of $\bar{K}[x,y]$ at $P$: $$ \bar{K}[C]_P = \{f/g \in \bar{K}(x,y) : g(P) \neq 0\} $$ $\bar{K}[C]_P$ has exactly one maximal ideal, namely $M_P \bar{K}[C]_P$. To prove this, simply observe that $M_P$ is a maximal ideal (essentially proved in Silverman, on p. 5 -- the quotient is $\bar{K}[C]_P / M_P \bar{K}[C]_P = \bar{K}$ which is a field), and every element of $\bar{K}[C]_P \setminus M_P$ is a unit, so there can be no other maximal ideals. By Proposition II.1.1 of Silverman (equivalently, Exercise 2.1 of Silverman), $\bar{K}[C]_P$ is a discrete valuation ring. (A discrete valuation ring is, by definition, a principal ideal domain with exactly one nonzero maximal ideal.)
From here, the theorem that you want is proved as follows:
Theorem: Let $R$ be a discrete valuation ring with maximal ideal $M$. Let $\pi \in M \setminus M^2$. Then $\pi$ is a generator of $M$ (i.e. $M = \pi R$).
Proof: By hypothesis, $R$ is a principal ideal domain, so $M$ is a principal ideal. Let $t$ be a generator of $M$. By a standard result in algebra, every principal ideal domain is a unique factorization domain. Hence $\pi = u\cdot t^k$ for some unit $u \in M$ and some non-negative integer $k$. If $k=0$, then $\pi = u$ is a unit, contradicting $\pi \in M$. If $k \geq 2$, then $\pi \in M^2$, contradicting $\pi \notin M^2$. Hence $k=1$ and $\pi = ut$, so $t = \pi u^{-1}$. Since $t$ is a multiple of $\pi$, every element of $M$ is a multiple of $\pi$. Furthermore, $\pi \in M$. Hence $M = \pi R$.