On page 45 the book states that for any $\lambda \in \mathbb C \setminus \{0\}$ and any star subalgebra $B$ of a $C^\ast$ algebra $A$ with $1_B \neq 1_A$, $b -\lambda 1_B$ is invertible in $B$ if and only if $b-\lambda 1_A$ is invertible in $A$. Therefore
$$ \sigma_A(b)= \sigma_B(b) \cup \{0\}$$
This should be $ \sigma_A(b) \cup \{0\}= \sigma_B(b) \cup \{0\}$, shouldn't it?
The version Murphy gives is correct too, and incorporates the fact that $0$ is an element of $\sigma_A(b)$ in this situation. If $b$ were invertible in $A$, then the C*-algebra generated by $b$ would automatically contain $1_A$. There was a question and answer about this fact here.
(Actually what is shown there is stronger, as it is not assumed that $B$ has an identity; subalgebras not containing the identity cannot contain invertible elements, and this is a stronger statement than is true for other kinds of algebras.)
In the case $1_A\notin B$ , we always have $0\in\sigma_A(b)$.
$\sigma_A(b)=\sigma_{B\oplus \mathbb{C}1_A}(b)$ since $B\oplus \mathbb{C}1_A$ is a colsed $*$-subalgebra of $A$, and we know $0\in\sigma_{B\oplus \mathbb{C}1_A}(b)$.
