Relation on partitions having $1$ as summand If $p_1(n)$ is the number of partitions of n having $1$ as a summand, then $p_1(n) = p(n − 1)$ for $n ≥ 2,$
for all such partitions of n can be obtained by putting $+1$ after each partition of $n − 1$. Thus for
$n ≥ 2, p(n) =$(number of partitions without $1$ as a summand)$ + p(n − 1)$
$5 = 5 = 2 + 3 \therefore p(5) = 2 + p(4) = 2 + 5 = 7$
$6 = 6 = 2 + 4 = 3 + 3 = 2 + 2 + 2 \therefore p(6) = 4 + p(5) = 4 + 7 = 11$
 A: $p_1(n)$ represents the number of partitions of $n$ such that for some number $m$, we have $n=m+1$.  Therefore, our problem has been reduced to the number of ways we can partition $m$. But $m=n-1$. Therefore, $p_1(n)=p(m)=p(n-1)$
Now, suppose that $n\geq 2$ and $p(n)$ is the number of ways we have partitioned $n$. 
$$\{\text{ways we can partition n}\}=$$
$$\{\text{ways we can partition $n$ without 1 as a summand}\} \cup \{\text{ways we can partition $n$ with 1 as a summand}\}$$
Since the two sets are disjoint, their numbers add up. Also, since the last set by definition is $p_1(n)$, and we already showed that $p_1(n)=p(n-1)$, you get the claim of the book.
In a more verbose fashion,
There are two possibilities when you partition $n$. Either you partition it in a way that a summand of $1$ occurs, or you partition it in a way that no summand of $1$ occurs. Since the two possibilities contradict each other, the numbers of ways that you can partition $n$ is equal to the sum of the number of ways for each possibility.
