Answer
Please see the work below.
Work Step by Step
We know that
$1.00g PA\times \frac{1 mol \space PA}{81.994\space g\space PA}\times \frac{2\space mol \space NaOH}{1 mol\space PA}\times \frac{40.00g NaOH}{1 \space mol \space NaOH}=0.976g\space NaOH$