= Solution
Write $q=1-p$, $\mu=q-p$, $\sigma=2\sqrt{pq}$, and $\phi_S(x)=\prod_{j\in S}(x_j-\mu)/\sigma$. The functions $\phi_S$ form the <p-biased product measure> <orthonormal basis>, so the <p-biased Fourier coefficient>[Fourier expansion] is $f=\sum_S\widehat f_p(S)\phi_S$. The normalized <discrete derivative of a Boolean function> satisfies
$$
D_i f=\sum_{S\ni i}\widehat f_p(S)\phi_{S\setminus\{i\}}.
$$
Applying <Parseval identity> and then exchanging two finite sums gives
$$
\operatorname{Inf}_i(f)=\lVert D_i f\rVert_2^2
=\sum_{S\ni i}\widehat f_p(S)^2,
\qquad
\mathbf I(f)=\sum_S|S|\widehat f_p(S)^2.
$$
The <noise operator on the Boolean hypercube> acts diagonally on the same basis: $T_\rho\phi_S=\rho^{|S|}\phi_S$. Hence the <noise stability> is
$$
\operatorname{Stab}_\rho(f)=\langle f,T_\rho f\rangle
=\sum_S\rho^{|S|}\widehat f_p(S)^2.
$$
Its <derivative> is
$$
\frac d{d\rho}\operatorname{Stab}_\rho(f)
=\sum_{S\ne\varnothing}|S|\rho^{|S|-1}\widehat f_p(S)^2.
$$
Taking the right-hand value at $\rho=0$ leaves exactly the <linear Fourier weight> $\sum_i\widehat f_p(\{i\})^2$, while taking the left-hand value at $\rho=1$ gives $\mathbf I(f)$.
Solved by gpt-5.6-sol high.
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