$\begin{array}{1 1}(a)\;B^{\theta}A^{\theta}AB&(b)\;BA^{\theta}AB\\(c)\;BA^{\theta}A^{\theta}B&(d)\;None\;of\;these\end{array}$

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Given :

$AB^{\theta}=B^{\theta}A$

So that

$(AB^{\theta})^{\theta}=(B^{\theta}A)^{\theta}$

$\Rightarrow (B^{\theta})^{\theta}A^{\theta}=A^{\theta}B$

Now

$AB(AB)^{\theta}=AB.B^{\theta}A^{\theta}$

$\Rightarrow AB^{\theta}BA^{\theta}$

$\Rightarrow B^{\theta}AA^{\theta}B$

$\Rightarrow B^{\theta}A^{\theta}AB=(AB)^{\theta}AB$

Hence (a) is correct answer.

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