The point that you are missing is that $P(f)$ as used in Eq.$(4.59)$ is no longer the minimum-bandwidth function that is nonzero only for $f \in [-W,W]$ but something that is nonzero for $f \in [-W(1+\alpha),W(1+\alpha)]$ where $0 < \alpha \leq 1$.
The Nyquist criterion for no intersymbol interference is not that $P(f)$ has constant value for $f \in [-W,W]$ but that if we add up frequency-shifted versions of $P(f)$ as in Eq.$(4.53)$, the sum is a constant for all $f$. Now, regardless of what $P(f)$ actually is (we don't care diddlysquat whether $P(f)$ is band-limited or not)
$$S(f) = \sum_{n=-\infty}^\infty P(f-nR_b) = \sum_{n=-\infty}^\infty P(f-n(2W))\tag{1}$$
is a periodic function of $f$ with period (not necessarily least period) $2W$. Disbelievers should carefully compute what $S(f+2W)$ is, and after murmuring a few shibboleths such as $-\infty+1 = -\infty$ and $\infty+1 = \infty$, satisfy themselves that $S(f+2W)$ is indeed equal to $S(f)$. That being said, Eq,$(4.53)$ is saying that $S(f)$ has constant value for all $f$, that is, $P(f)$ must be such that when we add up frequency-shifted copies of $P(f)$, the sum is a constant for all $f$, and in particular for all $f\in [W,W]$ which is one period of $S(f)$.
So, let's go back to Eq.$(4.59)$ where $P(f)$ has support $[-W(1+\alpha),W(1+\alpha)]$ with $0 < \alpha \leq 1$. Then, for $n \geq 2$, $P(f\pm n(2W))$ has support that is disjoint from $[-W,W]$. Thus, if we are looking only at $f\in [-W,W]$, we can cheerfully discard $P(f\pm n(2W))$ as being irrelevant in the sum $(1)$ since we are restricting attention to $f\in [-W,W]$. Eq.(4.59) is thus saying that in view of our restricted gaze, what we need is for $P(f)$ should be such that
$P(f) + P(f+2W) + P(f-2W)$ equals $\dfrac{1}{2W}$ for $f\in [-W,W]$. By periodicity of $S(f)$ in $(1)$, the sum in $(1)$ equals $\dfrac{1}{2W}$ for all $f$ too.
I'm unable to see how if the condition specified in the first image is true then that one in the second image becomes also true. I tried plugging W as $R_{b}/2$ and $R_{b}$ and then grouping terms to reach the condition but that didn't work and then after thinking for a bit it felt that the condition in the first image does not make sense. If $P(f)$ has bandwidth $W$ then the two other terms added to it will have no effect on it (each is $2W$ wide as well.)
