= Solution
At an interior <statistical parameter> $\theta$, <local asymptotic normality> means that there are random vectors $\Delta_{n,\theta}$ and a symmetric nonnegative <Fisher information matrix> $I_\theta$ such that, for each fixed $h\in\mathbb R^p$,
$$
\log\frac{\prod_{i=1}^n f(\theta+h/\sqrt n,Y_i)}{\prod_{i=1}^n f(\theta,Y_i)}
=h^T\Delta_{n,\theta}-\frac12h^TI_\theta h+o_{P_\theta^n}(1),
\qquad \Delta_{n,\theta}\xrightarrow{d}N_p(0,I_\theta).
$$
A common stronger definition requires this expansion also for $h_n\to h$. The fixed-$h$ version suffices here. A regular identifiable model usually has positive definite <Fisher information>; nonsingularity is unnecessary for the contiguity conclusion.
Here are sufficient differentiability conditions for a proof by <Taylor's theorem>. Suppose the local <probability density functions> have common support, $\ell_\vartheta(y)=\log f(\vartheta,y)$ is twice continuously differentiable near $\theta$, differentiation can pass through the normalization integral twice, the <score function> $s_\theta=\nabla\ell_\theta$ has finite second moment, and the local <Hessian matrix> is bounded in <operator norm> by an <integrable envelope of a function class>. These hypotheses imply
$$
\mathbb E_\theta s_\theta=0,\qquad
I_\theta=\mathbb E_\theta[s_\theta s_\theta^T]
=-\mathbb E_\theta\nabla^2\ell_\theta.
$$
The first identity differentiates $\int f_\theta=1$ once; the second uses $\nabla^2 f_\theta=f_\theta(\nabla^2\ell_\theta+s_\theta s_\theta^T)$ and differentiates twice. The <central limit theorem> gives $\Delta_{n,\theta}=n^{-1/2}\sum_i s_\theta(Y_i)\xrightarrow dN_p(0,I_\theta)$. Taylor's integral remainder gives
$$
\sum_i(\ell_{\theta+h/\sqrt n}(Y_i)-\ell_\theta(Y_i))
=h^T\Delta_{n,\theta}+
h^T\!\left[\frac1n\sum_i\int_0^1(1-t)\nabla^2\ell_{\theta+th/\sqrt n}(Y_i)\,dt\right]h.
$$
The <weak law of large numbers> and the <continuity> and integrable-envelope hypotheses make the bracket converge in probability to $-I_\theta/2$. Indeed, the <expectation> of the supremum of the <Hessian matrix> difference over a shrinking neighborhood tends to zero by <dominated convergence>, and <Markov inequality> controls its empirical average. This proves <local asymptotic normality> and explains the role of both <derivatives>.
Write $Q_n\triangleleft P_n$ for <contiguity of probability measures>: for every measurable sequence $A_n$, $P_n(A_n)\to0$ implies $Q_n(A_n)\to0$. <Mutual contiguity> requires this in both directions. A useful form of <Le Cam's first lemma> is the following: if the <Radon-Nikodym derivative> $L_n$ of the absolutely continuous part of $Q_n$ relative to $P_n$ converges in distribution under $P_n$ to $L$, with $L>0$ almost surely and $\mathbb EL=1$, then the two sequences are <mutually contiguous>. This includes the usual case $L_n=dQ_n/dP_n$. The mean-one condition ensures <uniform integrability> of $L_n$ and vanishing mass of any singular part; positivity ensures the reverse implication.
Take $P_n=P_\theta^n$ and $Q_n=P_{\theta+h/\sqrt n}^n$. Their <likelihood ratio> on the support of $P_n$ satisfies, by <local asymptotic normality> and the <continuous mapping theorem>,
$$
L_n\xrightarrow d L=\exp(W-v/2),\qquad
W\sim N(0,v),\quad v=h^TI_\theta h.
$$
This limit is positive and $\mathbb EL=e^{-v/2}e^{v/2}=1$, including $v=0$. The conditions of <Le Cam's first lemma> therefore hold, giving \b[<mutual contiguity> of the product sampling laws]. The products, rather than the one-observation laws, are the relevant sequences.
For any $\varepsilon>0$, let $A_n=\{\|\widehat\theta_n-\theta\|>\varepsilon\}$. <Consistency (statistics)> under $P_\theta^n$ makes $P_\theta^n(A_n)\to0$, and <contiguity> gives $Q_n(A_n)\to0$. Thus \b[the <estimator> remains consistent for $\theta$ under each fixed local alternative]. Since $\|\theta+h/\sqrt n-\theta\|\to0$, it is also consistent for the moving parameter in the sense that $\|\widehat\theta_n-(\theta+h/\sqrt n)\|\to0$ in probability under $Q_n$.
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