### Step-by-Step Solution

#### Part (1): Finding the Equation of $C$

Given the eccentricity of the ellipse is $\frac{\sqrt{2}}{2}$, we can write the eccentricity formula as:

$$\frac{\sqrt{a^2 - b^2}}{a} = \frac{\sqrt{2}}{2}$$

Squaring both sides, we get:

$$\frac{a^2 - b^2}{a^2} = \frac{1}{2}$$

Solving for $a^2$ and $b^2$, we find:

$$a^2 = 2b^2$$

Given that the circle with the diameter of the short axis of $C$ is tangent to the line $y = ax + 6$, the radius of this circle is $b$, and the distance from the origin to the line is $\frac{6}{\sqrt{a^2 + 1}}$, we equate this to $b$:

$$b = \frac{6}{\sqrt{a^2 + 1}}$$

Substituting $a^2 = 2b^2$ into the equation and solving, we find:

$$a^2 = 8, \quad b^2 = 4$$

Therefore, the equation of $C$ is:

$$\boxed{\frac{x^2}{8} + \frac{y^2}{4} = 1}$$

#### Part (2): Determining if $k \cdot k'$ is a Constant Value

Given $|AP| \cdot S_2 = |BP| \cdot S_1$, we analyze the ratios and find that:

$$\frac{|AP|}{|BP|} = \frac{S_1}{S_2} = \frac{\frac{1}{2}|AP||PQ|\sin\angle APQ}{\frac{1}{2}|BP||PQ|\sin\angle BPQ} = \frac{|AP|\sin\angle APQ}{|BP|\sin\angle BPQ}$$

This implies $\sin \angle APQ = \sin \angle BPQ$, and since $\angle APQ + \angle BPQ = \angle ABP \in (0, \pi)$, we have $\angle APQ = \angle BPQ$. This means $PQ$ bisects $\angle APB$, leading to $k_{AP} + k_{BP} = 0$.

Let $A(x_1, y_1)$, $B(x_2, y_2)$, and $P(x_0, y_0)$. From the system of equations:

$$\left\{ \begin{array}{l} \frac{x^2}{8} + \frac{y^2}{4} = 1 \\ y = k(x - 1) \end{array} \right.$$

We eliminate $y$ and find:

$$(1 + 2k^2)x^2 - 4k^2x + 2k^2 - 8 = 0$$

From this, we get:

$$x_1 + x_2 = \frac{4k^2}{1 + 2k^2}, \quad x_1x_2 = \frac{2k^2 - 8}{1 + 2k^2}$$

Given $k_{AP} + k_{BP} = 0$, we derive:

$$2k(2k^2 - 8) - 4k^2(y_0 + kx_0 + k) + 2x_0(y_0 + k)(1 + 2k^2) = 0$$

Simplifying, we find:

$$2y_0(x_0 - 1)k^2 + (x_0 - 8)k + x_0y_0 = 0$$

Given $P(x_0, y_0)$ lies on the ellipse, we have $x_0^2 + 2y_0^2 = 8$. This simplifies to:

$$(2y_0k - x_0)[(x_0 - 1)k - y_0] = 0$$

Since $P$ does not lie on the line $l: y = k(x - 1)$, we conclude:

$$2y_0k - x_0 = 0 \Rightarrow k \cdot \frac{y_0}{x_0} = k \cdot k' = \frac{1}{2}$$

Therefore, $k \cdot k'$ is a constant value, and:

$$\boxed{k \cdot k' = \frac{1}{2}}$$