To find $ p^2 + q^2 $ for the hyperbola $\mathrm{H}: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, given that the product of the focal distances from a point $ \mathbf{P}(4, 2\sqrt{3}) $ to the foci is 32, we follow these steps:

Verify the Point on the Hyperbola:

The point $ P(4, 2\sqrt{3}) $ satisfies the hyperbola equation:

$ \frac{16}{a^2} - \frac{12}{b^2} = 1 \quad \ldots \text{(i)} $

Product of Focal Distances:

The distances from $ P $ to the foci $ S $ and $ S' $ are:

$ SP = e\left(4 - \frac{a}{e}\right), \quad S'P = e\left(4 + \frac{a}{e}\right) $

Therefore, the product:

$ SP \cdot S'P = 16e^2 - a^2 = 32 $

Relating $ e $, $ a $, and $ b $:

Using the identity for eccentricity:

$ e^2 = 1 + \frac{b^2}{a^2} $

Substituting in the product equation gives:

$ 16\left(1 + \frac{b^2}{a^2}\right) - a^2 = 32 \quad \ldots \text{(ii)} $

Solving for $ a^2 $ and $ b^2 $:

From equations (i) and (ii), solve for $ a^2 $ and $ b^2 $:

$ a^2 = 8, \quad b^2 = 12 $

Calculate $ p^2 + q^2 $:

The conjugate axis length is $ p = 2b $ and the latus rectum is $ q = \frac{2b^2}{a} $. Thus:

$ p^2 + q^2 = (2b)^2 + \left(\frac{2b^2}{a}\right)^2 $

Substituting $ b^2 = 12 $ and $ a^2 = 8 $ gives:

$ p^2 + q^2 = 4b^2 + \frac{4b^4}{a^2} = 48 + \frac{4 \times 144}{8} $

$ p^2 + q^2 = 48 \cdot \left(1 + \frac{12}{8}\right) = 120 $

Therefore, $ p^2 + q^2 = 120 $.