For $A= (A_1,\cdots,A_d)\in {\cal L}(E)^d$ such that $A_iA_j=A_jA_i$ for all $i,j$. Why $$\sum_{f\in F(n,d)} A_{f}^*LA_{f}=\displaystyle\sum_{|\alpha|=n}\frac{n!}{\alpha!}{A^*}^{\alpha}LA^{\alpha},\,\;\forall\, L\in\mathcal{L}(E)?$$ Note that $F(n,d)$ denotes the set of all functions from $\{1,\cdots,n\}$ into $\{1,\cdots,d\}$ and $A_f:=A_{f(1)}\cdots A_{f(n)}$, for $f\in F(n,d)$. Also $\alpha = (\alpha_1, \alpha_2,...,\alpha_d) \in \mathbb{Z}_+^d;\;\alpha!: =\alpha_1!\cdots\alpha_d!,\;|\alpha|:=\displaystyle\sum_{j=1}^d|\alpha_j|$; $A^*=(A_1^*,\cdots,A_d^*)$ and $A^\alpha:=A_1^{\alpha_1} A_2^{\alpha_2}\cdots A_d^{\alpha_d}$.
I'm facing dificulties to understand the following proof:
Proof $$\sum_{f\in F(n,d)} A_{f}^*LA_{f}=\displaystyle\sum_{|\alpha|=n}{n\choose \alpha}\ {A^*}^{\alpha}LA^{\alpha}\ .$$ It is just an instance of the expansion of the $n$-power of the sum of $d$ commuting objects in a ring, $$(X_1+\dots X_d)^n=\sum_{\alpha\in\mathbb{N}^d\atop |a|=n} {n\choose \alpha}X^\alpha, $$ with $X:=X_1^{\alpha_1}X_2^{\alpha_2}\dots X_d^{\alpha_d}.$ So if $X_j$ is the linear operator on $\mathcal{L}(H)$ defined by $L\mapsto A_j^*LA_j$ , we get the desired formula.
Observe that $A_f=\prod_{k=1}^n A_{f(k)}$ and that $f(k)\in\{1,\ldots,d\}$.
Also, for some chosen $f$, set $\alpha_k:=|\{j\in\{1,\ldots,n\}:f(j)=k\}|$, where the bars $|\,|$ means "cardinality of the set". In other words: $\alpha_k=|f^{-1}(k)|$, the cardinality of the fiber of $k$ for some chosen $f$.
And because $A_{i,j}=A_{j,i}$ for all $i,j\in\{1,\ldots,n\}$ then $A_f=A^\alpha$ for some $\alpha\in\Bbb N_{\ge 0}^d$ such that $|\alpha|=n$. By example, if there is no $k\in\{1,\ldots,n\}$ such that $f(k)=d$ then $\alpha_d=0$, and if there are exactly a pair $i,j$ such that $f(i)=f(j)=7$ then $\alpha_7=2$, etc...
But for each $\alpha$ there are $\binom{n}{\alpha}$ distinct $f$ such that $A_f=A^\alpha$ because $A_{i,j}=A_{j,i}$ for all $i,j\in\{1,\ldots,n\}$, that is there are $\binom{n}\alpha$ different ways to order the list $f(1),f(2),\ldots, f(n)$ for some chosen $f$.
By example: set $n:=5$ and $d:=10$ and you want to know in how many distinct orders can be rearranged the list $1,\color{red}{3,3},\color{blue}5,\color{green}9$. From the multinomial theorem we knows that there are $\binom5{1,\color{red}2,\color{blue}1,\color{green}1}$ distinct orders for this list.
Now observe that $(a+b)^2=(a+b)(a+b)=a^2+ab+ba+b^2$. Then if $ab=ba$ (that is, if $a$ and $b$ belong to a commutative ring) then we can conclude that $(a+b)^2=a+2ab+b^2$.
This can be generalized as follows: let $A_i\in R$, where $(R,+,\cdot)$ is a commutative ring, then
$$\begin{align}(A_1+A_2+\ldots+A_d)^n&=\sum_{\substack{\alpha_1+\alpha_2+\ldots+\alpha_d=n\\\alpha_j\ge 0\,\forall j\in\{1,\ldots,d\}}}\frac{n!}{\alpha_1!\alpha_2!\cdots\alpha_d!}A_1^{\alpha_1}A_2^{\alpha_2}\cdots A_d^{\alpha_d}\\&=\sum_{\substack{\alpha\in{\Bbb N}_{\ge 0}^d\\|\alpha|=n}}\binom{n}\alpha A^\alpha\end{align}$$
for $\alpha:=(\alpha_1,\alpha_2,\ldots,\alpha_d)$, what means that there are $\binom{n}\alpha$ distinct ways to order the list given by
$$\underbrace{1,1\ldots,1}_{\alpha_1\text{ times}},\underbrace{2,\ldots,2}_{\alpha_2\text{ times}},\underbrace{3,\ldots,3}_{\alpha_3\text{ times}},\ldots,d-1,\underbrace{d,\ldots,d}_{\alpha_d\text{ times}}$$
where each $\alpha_j\ge 0$ and $\alpha_1+\alpha_2+\ldots+\alpha_d=n$. This is the reason why the proof uses the identity
$$(X_1+\dots+ X_d)^n=\sum_{\substack{\alpha\in\mathbb{N^d_{\ge0}}\\ |\alpha|=n}} {n\choose \alpha}X^\alpha$$
to try to show why the identity $$\sum_{f\in F(n,d)} A_{f}^*LA_{f}=\sum_{|\alpha|=n}{n\choose \alpha}\ {A^*}^{\alpha}LA^{\alpha}$$ holds.