Before tackling the question itself, it is perhaps useful to discuss a minor point regarding the fact that the unit of
$p A p$ is $p$, rather than $I$. To highlight this difference,
whenever we are given an element
$b\in p A p$,
we will write $\sigma _p(b)$ for the spectrum of $b$ relative to
$p A p$,
reserving the notation
$\sigma (a)$
for the spectrum of any element $a\in A$ relative to $A$ (or, equivalently, to $B(H)$).
Leaving aside the trivial case in which $p=1$, observe that no element $b\in pAp$ is invertible relative to $A$, so 0 is always in $\sigma (b)$. In
fact it is easy to show that, for every such $b$, one has
$$
\sigma (b) = \sigma _p(b)\cup \{0\}.
\tag{*}
$$
Likewise, if $b\in p A p$, and $f$ is a holomorphic function on a neighborhood of $\sigma _p(b)$, we will denote by
$f_p(b)$
the outcome of the holomorphic functional calculus computed relative to $pAp$. As before, we will reserve the
undecorated expression $f(a)$ for the holomorphic functional calculus relative to $A$.
In the event that $f$ is holomorphic on the larger set $\sigma _p(b)\cup \{0\}$, one may easily prove that
$$
f(b)=f_p(b)+f(0)(1-p),
\tag{**}
$$
for every $b\in p A p$.
This said, let $b\in \mathscr A$ and let $f$ be a holomorphic function on an open set $U$ such that $\sigma _p(b)\subseteq U$.
CASE 1: Assuming first that $0\in U$, we have by ($*$)
that $f$ is
also holomorphic on a neighborhood of $\sigma (b)$ and we have by
hypothesis that $f(b)\in \mathscr A$. Applying ($**$) it then follows that
$$
f_p(b)=pf(b)p \in p\mathscr A p,
$$
as desired.
CASE 2: $0\notin U$.
In this case it is clear that $0\notin \sigma _p(b)$, and
since $\sigma _p(b)$ is compact, one has that $r:=\text{dist}(0,\sigma _p(b))>0$. So
$$
\sigma _p(b)\subseteq V:= \mathbb C\setminus \overline {B_{r/2}(0)}.
$$
By restricting $f$ to $U\cap V$, we may assume that $U\subseteq V$. The open ball $B_{r/3}(0)$ is therefore disjoint from
$U$, so we may extend $f$ to $U\cup B_{r/3}(0)$ by declaring it to be identically zero on $B_{r/3}(0)$. Consequently the extended $f$
is now defined on a neighborhood of $\sigma _p(b)\cup \{0\}$.
Since the extension process did not change the values of $f$ on points of $\sigma _p(b)$, the outcome of $f_p(b)$ remains
unchanged and hence the conclusion follows as in case 1.
Best Answer
This is Proposition 2.7.19 in Willett and Yu's book "Higher Index Theory" (you can find a copy of a draft of the book here, or from Rufus Willett's homepage). The idea of the proof is to reduce to the separable case, and appeal to the following result of Brown:
Using this, we define a map $\phi:A\otimes\mathbb K\to pAp\otimes\mathbb K$ by $\phi(a)=vav^*$. This is a $*$-isomorphism, and its composition with the inclusion $pAp\otimes\mathbb K\hookrightarrow A\otimes\mathbb K$ yields an isomorphism on $K$-theory, so in particular the inclusion (after tensoring with $\mathbb K$) induces an isomorphism on $K$-theory. Finally, we have a commutative diagram \begin{align*} \begin{array}{ccc} pAp & \hookrightarrow & A \\ \downarrow & & \downarrow \\ pAp\otimes\mathbb K & \hookrightarrow & A\otimes\mathbb K \end{array} \end{align*} where the vertical arrows are "upper left corner" inclusions. The vertical maps and the bottom map induce isomorphisms on $K$-theory, hence so does the top map.