1. Let $z = r e^{i\alpha}, \quad w = s e^{i\beta}$ where $r = |z|$, $s = |w|$, $\alpha = \arg(z)$, and $\beta = \arg(w)$.

2. Given: $|zw| = 1$ But $|zw| = |z|\,|w| = rs$ so $rs = 1.$

3. Also given: $\arg(z) - \arg(w) = \frac{\pi}{2}$ i.e. $\alpha - \beta = \frac{\pi}{2}.$

4. Now compute $z\overline w$:

   = rs\, e^{i(\alpha-\beta)}.$$ Using $rs=1$ and $\alpha-\beta=\frac{\pi}{2}$, $$z\overline w = e^{i\pi/2} = i.$$

5. Next compute $\overline z w$:

   = rs\, e^{i(\beta-\alpha)}.$$ Again using $rs=1$ and $\beta-\alpha=-\frac{\pi}{2}$, $$\overline z w = e^{-i\pi/2} = -i.$$

6. Now check the options:

   • A: $z\overline w = \dfrac{1-i}{\sqrt2}$ → false, since $z\overline w = i$
   • B: $\overline z w = i$ → false, since $\overline z w = -i$
   • C: $z\overline w = \dfrac{-1+i}{\sqrt2}$ → false, since $z\overline w = i$
   • D: $\overline z w = -i$ → true

Therefore, the correct option is D.