Given the density is
$$f(x; a, \theta) := \theta a^{\theta} x^{-(\theta+1)}\boldsymbol 1_{(a,\infty)}(x).\tag 1\label 1$$
The pdf of the first order statistic $X_{(1)}$ can be easily shown to be $$g(x_1):= n\theta a^{n\theta}{x_1}^{-(n\theta+1)} \boldsymbol 1_{(a,\infty)}(x_1)\tag 2\label 2.$$
Observation $1.$ For known $\theta,~\hat a:= X_{(1)} $ is sufficient and complete for $a.$
Use $\eqref 2$ to express the joint pdf of the sample in the form $$f(\mathbf x) = g\left(\hat a; a\right)\times \left\{\frac1n \theta ^{n-1}\left(\prod x_i\right)^{-(\theta+1)}{\hat a}^{n\theta+1}\right\} .\tag 3$$ Sufficiency follows from Neyman factorization theorem.
Consider an arbitrary Borel measurable function $\varphi$ such that $$\mathbb E\left[\varphi\left(\hat a\right)\right] = 0, ~\forall a\in (0,\infty).\tag 4\label 4$$
$\eqref 4$ implies $$\int_a^\infty\varphi\left(\hat a\right){\hat a}^{-(n\theta+1)}~\mathrm d\hat a = 0, ~\forall a\in (0,\infty).\tag 5\label 5$$ Define (denoting $\varphi:= \varphi^+-\varphi^-$) $\nu^{\pm}(A) :=\int_A\varphi^{\pm}\left(\hat a\right){\hat a}^{-(n\theta+1)}~\mathrm d\hat a; ~~A:= [a, \infty); $ from $\eqref 5, ~\nu^+(A) = \nu^-(A)$ whence $\varphi^+ = \varphi^- ~\textrm{a.s.} ~[\lambda].$ The result follows.
$\blacksquare$
Observation $2.$ Define $Z:= \sum_{i=1}^n\ln\frac{Y_i}{Y_1} $ where $Y_i$ are the order statistics. Then $\hat \theta:= Z$ and $\hat a = Y_1$ are stochastically independent.
The approach would be to show $M_Z(t),$ the moment generating function of $Z$ doesn't depend on $a,$ whence by Observation $1.$ and Basu's theorem, $Z$ would be independent of $Y_1.$
$$M_Z(t) = n!\int_a^\infty\int_a^{y_n}\cdots\int_a^{y_2}\exp\left(t\sum_{i=1}^n\ln\frac{Y_i}{Y_1}\right)\theta^na^{n\theta}\prod_{i=1}^n y_1^{-(\theta+1)}~\mathrm dy_i ;$$
substitute $y_i\mapsto \frac a{y_i}~\forall i\in\{1,2,\ldots, n\}.$ It is easy to see it is one-to-one and $|\mathcal J| = a^n.$ Therefore, $M_Z(t)$ reduces to a form that doesn't depend on $a.$
$\blacksquare$
Now, we would concentrate on the distribution of $Z.$ Note that $\sum_{i=1}^n\ln\frac{X_i}{X_1}$ doesn't depend on the ordering of $x_2, x_3, \ldots, x_n.$ So assuming $x_1< x_2, \ldots, x_n, ~~\sum_{i=1}^n\ln\frac{X_i}{X_1} = \sum_{i=1}^n\ln\frac{Y_i}{Y_1}.$ Also
$$g(x_2, \ldots, x_n\mid x_1) = \frac{\prod_{i=2}^nf(x_i)}{[1- F(x_1)]^{n-1}} .\tag 6$$ Therefore, the characteristic function of $Z$ given $X_1 = x_1$
\begin{align}\phi(t) &= \mathbb E\left[\exp\left(it\sum_{i=1}^n\ln\frac{X_i}{X_1}\right)\bigg \vert ~x_1\right]\\ &=\left[\int_{x_2}^\infty\frac{\exp\left(it\ln\frac{x_2}{x_1}\right)f(x_2)}{1- F(x_1)}~\mathrm dx_2\right]^{n-1} \tag 7.\end{align}
From this, the pdf of $Z$ is (by Observation $2.$)
\begin{align}f(z) &= \frac1{2\pi}\int_{-\infty}^{\infty}\exp{(-itz)}\phi(t)~\mathrm dt.\end{align}
Simplify it to deduce (cf. $\rm [II]$)
$$f(z) = \frac{\theta}{\Gamma{(n-1)}}z^{n-2}e^{-\theta z}.\tag 8 \label 8$$
Observation $3.$$\hat \theta$ is complete.
For arbitrary Borel measurable function $\varphi,$
\begin{align}\mathbb E\left[\varphi\left(\hat \theta\right)\right] &= 0, ~\forall \theta\in (0, \infty)\\\implies \int_0^\infty \varphi\left(\hat \theta\right){\hat\theta}^{n-2}\exp{\left(-\theta\hat\theta\right)}~\mathrm d\hat\theta &= 0; \tag 9\label 9\end{align}
$\eqref 9$ resembles the kernel of a one-dimensional exponential family. So, $\hat \theta$ is complete for $\theta.$
$\blacksquare$
Theorem $1.$ The family $\{F\left(\hat\theta, \hat a;\theta, a\right): (\theta, a)\in \mathbb R_+\times \mathbb R_+\}$ is complete.
From $\eqref 2, \eqref 8$ and by Observation $2.,$ write down the density $f\left(\hat\theta, \hat a;\theta, a\right).$
For an arbitrary Borel measurable function $\varphi\left(\hat\theta, \hat a\right)$
\begin{align}\mathbb E\left[\varphi\left(\hat\theta, \hat a\right)\right] &= 0, ~\forall (\theta, a)\in \mathbb R_+\times \mathbb R_+\\ \implies \int_0^\infty\int_k^\infty f\left(\hat\theta, \hat a;\theta, a\right)\varphi\left(\hat\theta, \hat a\right)~\mathrm d\hat\theta\mathrm d\hat a & = 0\\ \overset{\textrm{Fubini}}{\implies} \int_k^\infty n\theta a^{n\theta}{\hat a}^{-(n\theta+1)}\left[\underbrace{\int_0^\infty\varphi\left(\hat\theta, \hat a\right) \frac{\theta}{\Gamma{(n-1)}}{\hat\theta}^{n-2}e^{-\theta \hat\theta}~\mathrm d\hat\theta}_{:= h\left(\hat a, \theta\right)}\right] ~\mathrm d\hat a & =0 ;\tag{ 10}\label{ 10}\end{align}
by Observation $1.$ and $\eqref{ 10}, ~\lambda \{\hat a: h\left(\hat a, \theta\right) \ne 0\} = 0, ~\forall \theta > 0.$ This means $\lambda\otimes\lambda\{\left(\theta,\hat a\right): h\left(\hat a, \theta\right) \ne 0\} = 0 $ whence $\lambda\{\left(\theta,\hat a\right): h\left(\hat a, \theta\right) \ne 0\} = 0, ~\forall \hat a~\textrm{a.s.}.$ By continuity of $h\left(\hat a, \cdot\right), ~h\left(\hat a, \cdot\right) = 0,~\forall \hat a~\textrm{a.s.}\overset{\textrm{Obs.}~3.}\implies \lambda\left\{\hat a: \varphi\left(\hat\theta,\hat a\right) \ne 0\right\} = 0, ~\forall \hat a~\textrm{a.s.}\implies \lambda\otimes\lambda\left\{\left(\hat\theta,\hat a\right): \varphi\left(\hat\theta,\hat a\right) \ne 0\right\} = 0.$
$\blacksquare$
References:
$\rm [I]$ Best Unbiased Estimators for the Parameters of a
Two-Parameter Pareto Distribution, S.K. Saksena, A.M. Johnson, Metrika, Volume $31,~ 1984,$ page $77-83.$
$\rm[II] $ Estimation of the Parameters
of the Pareto Distribution, H. J. Malik, Skandinavisk
Aktuarietidskrift $49, ~1966, ~144-157.$
