Consider a single qubit CPTP map $\mathcal{N}$ such that $$\mathcal{N}(I) = I + pZ,~~~~~~\mathcal{N}(Z) = (1-p)Z,$$
where $I$ and $Z$ are Pauli operators. For an $n$ qubit Pauli operator $P$, made only of $I$ and $Z$, I am trying to find $\mathcal{N}^{\otimes n}(P).$
Let the weight of $P$ be $|P|$. I know that $\mathcal{N}^{\otimes n}(P)$ will be multiplied by a prefactor of $(1-p)^{|P|}$. But how do I write the rest of the terms, without cluttering the notation too much?
I'm interpreting this as a notation question for $\mathcal{N}$ restricted to act on $\{I, Z\}$. One notation I like is to write $\mathbf{Z}^{\mathbf{b}}:= \bigotimes_{i=1}^k Z_i^{b_i}$, where $\mathbf{b}\in\{0,1\}^k$ is basically a binary mask for where $Z$ terms appear in $P$. $^1$
What happens to the $Z$ terms in $P$ is trivial, so lets just consider $P = I^{\otimes k}$ with $k < n$. We get something like a binomial expansion : \begin{align} \mathcal{N}^{\otimes k}(I^{\otimes k}) &= \bigotimes_{i=1}^k (I + pZ) \tag{1} \\&= I^{\otimes k} + p (ZII\cdots + IZI\cdots +IIZ\cdots + \cdots) \\&\qquad +p^2(ZZI\cdots + ZIZ \cdots + IZZ \cdots + \cdots) + \cdots \tag{2} \\&= I^{\otimes k} + p\sum_{\mathbf{b}:\, |\mathbf{b}|=1} \mathbf{Z}^{\mathbf{b}} + p^2 \sum_{\mathbf{b}:\, |\mathbf{b}|=2}\mathbf{Z}^{\mathbf{b}} + \cdots \tag{3} \\&= \sum_{w=0}^k p^w \sum_{\mathbf{b}:\, |\mathbf{b}|=w}\mathbf{Z}^{\mathbf{b}} \tag{4} \end{align} where each inner sum is over binary strings $\mathbf{b}\in \{0,1\}^k$, and $|\mathbf{b}| = \sum_i b_i$ denotes the weight of the bit string.
I don't really know how to apply the above notation to the general case that results in something both concise and easy to understand. Here was my attempt anyway:
Let $S \subseteq \{1, 2, \dots, n\}$ with $|S|=(n-k)$ specify all qubits acted on by $Z$ in $P$, i.e. there are $k$ many $I$'s in the Pauli string $P = \mathbf{Z}^{\mathbf{b}}$ and $|\mathbf{b}| = n-k$. Then we want a simple notation for the image of $k$-many $I$ terms in $P$ that result in something like Eq. (4), but interleaved with the $(n-k)$-many $Z$ terms also appearing in $\mathcal{N}(P)$ acting on all qubits in $S$. It's sufficient to notate "the set of all length-$n$ binary strings of weight $k+w$ containing 1's in the support of $S$ ", and then we just draw $\mathbf{b}$ from that set in the inner sums of Eq. (4). Define $$S(\{0,1\}^n) := \{x \in \{0,1\}^n: x_i = 1 \, \forall\, i \in S\}$$ and you would then have \begin{equation} \mathcal{N}^{\otimes k}(P) = (1-p)^{n-k} \sum_{w=0}^k p^w \sum_{\mathbf{s} \in S(\{0,1\}^n):\\ |\mathbf{s}|=w +n-k}\mathbf{Z}^{\mathbf{s}} . \end{equation} This seems quite concise, but might be confusing to a reader.
$^1$ An alternative option to summing over weights of bitstrings is to sum over the relevant permutation group, e.g. see this answer, but this hides roughly the same complexity.
