Since the new model of watch A is more expensive than the old model of watch A and the new model of watch B is cheaper than the old model of watch B, but the difference in price is still $p\%$, the old model of watch A must be $p\%$ cheaper than the old model of watch B, and the new model of watch A must be $p\%$ more expensive than the new model of watch B.

Let the price of the old model of watch B be $x$. Then the price of the new model of watch B is $(1 - \frac{q}{100})x$. The old model of watch A then costs $(1 - \frac{p}{100})x$, while the new model of watch A costs $(1 + \frac{q}{100})(1 - \frac{p}{100})x$. Since from the previous paragraph the price of the new model of watch A is $(1 + \frac{p}{100})(1 - \frac{q}{100})x$, we have
$$\left(1 + \frac{q}{100}\right) \left(1 - \frac{p}{100}\right) x = \left(1 + \frac{p}{100}\right) \left(1 - \frac{q}{100}\right) x.$$
After cancelling $x$ from both sides, opening the parentheses and simplifying, we get $p = q$.

a. Since $p = q$, the prices of the new model of watch B and of the old model of watch A are both $(1 - \frac{p}{100})x$.

b. Since $p = q$, the price of the new model of watch A is $(1 - \frac{p^2}{10000})x$ which is less than the price of the old model of watch B.