The correct answer is $\boxed{\left( {\frac{{100 – {{\text{p}}_1}}}{{100 – {{\text{p}}_2}}}} \right){{\text{V}}_1}}$.
Let $V_1$ be the initial volume of the sludge, and let $V_2$ be the volume of the sludge after the moisture content has been reduced from $p_1\%$ to $p_2\%$. The mass of the sludge is constant, so we can write
$$V_1 \cdot (100 – p_1) = V_2 \cdot (100 – p_2)$$
Dividing both sides by $V_1$, we get
$$(100 – p_1) = \frac{V_2}{V_1} \cdot (100 – p_2)$$
$$\frac{100 – p_1}{100 – p_2} = \frac{V_2}{V_1}$$
$$V_2 = V_1 \cdot \frac{100 – p_1}{100 – p_2}$$
Therefore, the volume of the sludge after the moisture content has been reduced is given by
$$V_2 = \left( {\frac{{100 – {{\text{p}}_1}}}{{100 – {{\text{p}}_2}}}} \right){{\text{V}}_1}$$
Option A is incorrect because it divides by $100 + p_1$ instead of $100 – p_2$. Option B is incorrect because it divides by $100 – p_1$ instead of $100 + p_2$. Option C is incorrect because it divides by $100 + p_2$ instead of $100 – p_1$.