A question about ligand binding
You have calculated correctly for parts $\bf{a}$ and $\bf{b}$. However, to answer part $\bf{c}$, you need to use the given plot, the interception and slope of which can be calculated as follows:

The slope is $\tan \alpha = - \frac{1}{K_d}$:
$$\tan \alpha = \frac{1-0.2}{0.1-0.8} = -1.14 \ \Rightarrow \ \therefore \ \frac{1}{K_d} = 1.14$$
According to the given Scatchard equation (where  each receptor has a single ligand binding site), interception of the plot is also equal to $\frac{1}{K_d} = 1.14$. Therefore, the Scatchard equation can be rewritten as:
$$\frac{r}{[A]} = 1.14(1-r) \tag1$$
where $[A]$ is actually free ligand concentration (Wikipedia). Therefore, when $[A]= \pu{3.2 \mu M}$, find the relevant $r$ using the equation $(1)$, which is the average amount of $A$ per $\pu{mol}$ of $P$ $(r = \pu{0.785 \mu M})$.
This means: Total number of binding site per receptor is $1$ and only $0.789$ is bound by $A$. Protein concentration in the dialysis tube $= \pu{2.0 \mu M}$
Therefore, $A$ bound concentration of $P$ $ = \pu{2.0 \mu M} \times 0.785 = \pu{1.57 \mu M}$