For any point $(x_0, y_0)$ on parabola $y = ax^2 - 3x + 3$ ($a \neq 0$), there is
$$y_0 = a x_0^2 - 3x_0 + 3. \qquad \textcircled{1}$$
Suppose the symmetric point of $(x_0, y_0)$ with respect to line $y = x + m$ is $(x_1, y_1)$.
By $\frac{y_1 + y_0}{2} = \frac{x_1 + x_0}{2} + m$, $x_1 + y_1 = x_0 + y_0$, it follows that
$$x_1 = y_0 - m, \quad y_1 = x_0 + m.$$
Since $(x_1, y_1)$ lies on parabola $y^2 = 2px$, there is $(x_0 + m)^2 = 2p(y_0 - m)$. This is equivalent to
$$y_0 = \frac{1}{2p} x_0^2 + \frac{m}{p} x_0 + \frac{m^2}{2p} + m. \qquad \textcircled{2}$$
Due to the arbitrary taking of $(x_0, y_0)$, comparing (1) and (2) gives
$$\frac{1}{2p} = a, \quad \frac{m}{p} = -3, \quad \frac{m^2}{2p} + m = 3.$$
Therefore, we have $3 = \frac{m}{p} \cdot \frac{m}{2} + m = -3 \cdot \frac{m}{2} + m = -\frac{m}{2}$, and its solution is $m = -6$. Hence, in order, we can get $p = 2, a = \frac{1}{4}$, satisfying $a \neq 0$ and $p > 0$. Consequently, $apm = 2 \cdot \frac{1}{4} \cdot (-6) = -3$. $\square$