The answer is $\mathbf{(C)}$. Let us consider the four factors of the least common multiple $A(x) = (x-1)^2$, $B(x) = (x-2)^3$, $C(x) = x-3$ and $D(x) = x+1$. In order for the least common multiple of $p(x)$ and $q(x)$ to be $A(x) B(x) C(x) D(x)$, it is necessary that each of the four factors divides at least one of the two polynomials; on the other hand, each factor divides exactly one of the two polynomials, otherwise it would divide the greatest common divisor of the two (which we know to be equal to $(x-1)(x-2)$).

Let us first ignore the condition on the degree of the two polynomials. Once we have decided which of the four factors appear in $p(x)$, it is determined: it is the least common multiple between the product of the chosen factors and $(x-1)(x-2)$; likewise, $q(x)$ is also determined: it is the least common multiple between the factors not chosen and $(x-1)(x-2)$.

Note that $p(x)$ and $q(x)$ cannot have the same degree, since their product turns out to be $A(x) B(x) C(x) D(x) (x-1)(x-2)$, which has degree $9$, hence odd. We conclude that the number of ordered pairs of polynomials $(p(x), q(x))$ with the given greatest common divisor and least common multiple is $2^4 = 16$ (the number of subsets of the set of the four factors $A(x), B(x), C(x), D(x)$).

On the other hand, swapping the two polynomials of the pair reverses the inequality (which we know to be strict) between the degrees: exactly one pair out of every two is such that the degree of $p(x)$ is greater than that of $q(x)$. The correct answer is therefore $8$.