Step 1: Write the dissolving equation.
$\text{PQ}_2$ dissolves as $\text{PQ}_2 \rightarrow \text{P}^{2+} + 2\text{Q}^-$. If its solubility is $s$, then $[\text{P}^{2+}] = s$ and $[\text{Q}^-] = 2s$. The solubility product is $K_{sp} = [\text{P}^{2+}][\text{Q}^-]^2 = s(2s)^2 = 4s^3$.
Step 2: Find the original solubility.
Set $4s^3 = 4 \times 10^{-12}$, so $s^3 = 10^{-12}$ and $s = 10^{-4}$ mol per litre. This is the solubility before adding any RQ.
Step 3: Find the new solubility.
The solubility drops by a factor of 100, so the new solubility is $s' = \frac{10^{-4}}{100} = 10^{-6}$ mol per litre. Now $[\text{P}^{2+}] = s' = 10^{-6}$.
Step 4: Use the common ion from RQ.
RQ is highly soluble and provides extra $\text{Q}^-$ ions. Let the added RQ give a concentration $x$ of $\text{Q}^-$. Since the salt barely dissolves now, the $\text{Q}^-$ from $\text{PQ}_2$ is tiny, so the total $[\text{Q}^-] \approx x$.
Step 5: Plug into the $K_{sp}$ expression.
The $K_{sp}$ must still hold: $K_{sp} = [\text{P}^{2+}][\text{Q}^-]^2$, so $4 \times 10^{-12} = (10^{-6})(x)^2$. Then $x^2 = \frac{4 \times 10^{-12}}{10^{-6}} = 4 \times 10^{-6}$, giving $x = 2 \times 10^{-3}$ mol per litre.
Step 6: Convert to millimoles per litre.
$x = 2 \times 10^{-3}$ mol per litre equals 2 millimoles per litre, which is the concentration of added RQ.
\[ \boxed{[\text{RQ}] \approx 2\ \text{mmol L}^{-1}} \]