Lemma 1: Let $R$ be a Noetherian ring and $(a) \subset I\in \mathrm{Spec(R)}$ such that $\mathrm{ht}(I) = 0$. Then $a$ is a
divisor of $0$.
Proof: Consider the localized ring $R_{I}$. Clearly, $R_{I}$ is a Noetherian ring and $\mathrm{Spec}(R_{I}) = \{I':= I_I\}$ (since $I$ has height $0$ and there is a bijection between $\mathrm{Spec}(R_I)$ and $\{A\in \mathrm{Spec}(R);\ A\cap (R\setminus I) = \emptyset\}$) we can conclude that $R_I$ is an Artinian ring. Theorefore $I'$ is nilpotent. Thus, there exists $s\in R\setminus I$ and $n\in\mathbb{R}$, such that $s a^{n} = 0$. Implying that $a$ is a $0$ divisor.
Suppose by reductio ad absurdum that $\mathrm{ht}(P/(a)) = \mathrm{ht}(P) = n$. Then there there exists a saturated chain of prime ideals in $R/(a)$ such that
$$Q_0 \subsetneq Q_1 \subsetneq \ldots\subsetneq Q_{n-1}\subsetneq P/(a). $$
It is well known that there is a bijection between the set $\mathrm{Spec}(R/(a))$ and $\{I\in \mathrm{Spec}(R); (a)\subset I\}$. Then $\forall i\in\{0,1,\ldots,n-1\}$, there exists a unique $P_i\in \mathrm{Spec}(R)$, such that $(a)\subset P_i$ and $P_i/(a) = Q_i$. Thus, we can define the following chain of prime ideals in $R$
$$P_0 \subsetneq P_1 \subsetneq \ldots\subsetneq P_{n-1}\subset P. $$
Since $\mathrm{ht}(P)=n$ the above chain is saturated. So $\mathrm{ht}(P_0) = 0$ and $(a)\subset P_0$. Using Lemma 1, $a$ is a zero divisor, which is a contradiction. Therefore $\mathrm{ht}(P/(a))<\mathrm{ht}(P)$.
Corollary 15.15, of the same book, states that the following inequality holds
$$\mathrm{ht}(P/(a)) \leq \mathrm{ht}(P)\leq \mathrm{ht}(P/(a)) +1, $$
so,
$$\left|\mathrm{ht}(P) - \mathrm{ht}(P/(a)) \right| \leq 1. $$
We already have proved that $ 0<\left|\mathrm{ht}(P) - \mathrm{ht}(P/(a)) \right|$. Therefore $$\mathrm{ht}(P) = \mathrm{ht}(P/(a)) +1.$$
$$\hspace{10cm}\square $$
