I am using this definition of PDO on page 19.
I want to verify my understanding by the example of exetior derivative.
The exterior derivative $d:\Omega^p(M) \rightarrow \Omega^{p+1}(M)$ can locally be given by, for a $p$-form, $$ w = \sum_{|I|=p} a_I dx_I \mapsto \sum_{|I|=p} \sum_{j=1}^n \frac{\partial{a_I}}{\partial x_j} dx_j \wedge dx_I$$
where $I$ subset of $\{0,\ldots, n \}$, $n = \dim M$.
Claim: $d$ is a linear PDO of order $1$.
The definition of PDO requiries that locally $P$ looks like $$f \mapsto \sum _{|\alpha| \le k } A^\alpha (y) \frac{\partial^{\alpha}}{\partial x_\alpha} f(y).$$
So I think it is not obvious what $A^\alpha$s are where $\alpha = (0,\ldots, 1, \ldots, 0)$, $1$ in the $j$th position.
In these coordinates. $\frac{\partial w}{\partial x_j}$ is $\sum_{|I|=p} \frac{\partial a_I}{\partial x_j} dx_I$
Depending on $n$, $A^\alpha$ is a matrix with values in $\pm1$, $0$.
Is my line of thought correct?
Your thoughts look correct. I'll just (attempt) to set the record straight.
In local coordinates $x_1,\ldots,x_n$, we can write $$d\left(\sum_{|I|=p}a_Idx_I\right)=\sum_{j=1}^ndx_j\wedge\left[\frac{\partial}{\partial x_j}\sum_{|I|=p}a_Idx_I\right]=\sum_{j=1}^nA^j\frac{\partial}{\partial x_j}\left(\sum_{|I|=p}a_Idx_I\right) .$$ We have $n$ different $\binom{n}{p+1}\times\binom{n}{p}$ matrices $A^1,\ldots A^n$, which we can conveniently index by $p-$ and $(p+1)-$element subsets of $\{1,\ldots,n\}$. For $I,J\subset\{1,\ldots,n\}$ with $|I|=p$, $|J|=p+1$, the $(J,I)$-entry of $A^j$ will be $\pm 1$ if $J=I\cup\{j\}$ (with the sign of the entry depending on $I$ in the obvious way), and $0$ otherwise.