Integrate the gradient function to find the equation of the curve.

The gradient of the curve is given by $\frac{dy}{dx} = 2px + 3$.
To find the equation of the curve, $y(x)$, we integrate the gradient with respect to $x$:

$$y = \int (2px + 3)\,dx$$ $$y = p\frac{x^2}{2/2} + 3x + C$$ $$y = px^2 + 3x + C$$

where $C$ is the constant of integration.

The curve passes through the point $(-2, 0)$, so substitute $x=-2$ and $y=0$ into the curve equation:

$$0 = p(-2)^2 + 3(-2) + C$$ $$0 = 4p - 6 + C$$ $$C = 6 - 4p \quad \text{(Equation 1)}$$

Assumption: The second point is $(\frac{1}{2}, 0)$. The handwritten text appears to show $\frac{1}{x}$, but this would make the x-coordinate dependent on $x$, which is not standard for a fixed point. We proceed with the interpretation of $(\frac{1}{2}, 0)$.

Substitute $x=\frac{1}{2}$ and $y=0$ into the curve equation:

$$0 = p\left(\frac{1}{2}\right)^2 + 3\left(\frac{1}{2}\right) + C$$ $$0 = p\left(\frac{1}{4}\right) + \frac{3}{2} + C$$ $$0 = \frac{p}{4} + \frac{3}{2} + C \quad \text{(Equation 2)}$$

Substitute Equation 1 ($C = 6 - 4p$) into Equation 2:

$$0 = \frac{p}{4} + \frac{3}{2} + (6 - 4p)$$

Combine the constant terms and $p$ terms:

$$0 = \frac{p}{4} - 4p + \frac{3}{2} + 6$$ $$0 = \frac{p}{4} - \frac{16p}{4} + \frac{3}{2} + \frac{12}{2}$$ $$0 = -\frac{15p}{4} + \frac{15}{2}$$

To solve for $p$, move the term with $p$ to the other side:

$$\frac{15p}{4} = \frac{15}{2}$$

Multiply both sides by 4:

$$15p = \frac{15}{2} \times 4$$ $$15p = 15 \times 2$$ $$15p = 30$$

Divide by 15:

$$p = \frac{30}{15}$$ $$p = 2$$

The value of $p$ is $\boxed{2}$.