2. For $d=1$ we have $w_0\colon=w_\emptyset=1$, $w_1(\o)\colon=w_{\{1\}}(\o)=\o$ and thus $P_1w_0=w_0$, $P_1w_1=-w_1$. Now for $A\sbe\{1,\ldots,d\}$ we have $w_A=w_{r_1}\otimes w_{r_2}\otimes\cdots\otimes w_{r_d}$, where $r_j=1$ if $j\in A$ and otherwise $r_j=0$. Since
$$
dP
=P_1\otimes1\otimes\cdots\otimes1
+1\otimes P_1\otimes1\otimes\cdots\otimes1+\cdots
+1\otimes\cdots\otimes1\otimes P_1
$$
we get
$$
dPw_A
=P_1w_{r_1}\otimes w_{r_2}\otimes\cdots\otimes w_{r_d}
+\cdots
+w_{r_1}\otimes\cdots\otimes P_1w_{r_d}
=|\{j:r_j=0\}|w_A-|\{j:r_j=1\}|w_A
=(|A^c|-|A|)w_A~.
$$