Answer (B): There are two cases, depending on whether the denominator of the fraction is constant.

CASE 1. Suppose that $c = 0$. Then the graph of $y = \frac{ax+b}{d}$ is a line. It is symmetric with respect to the line $y = x$ if the slope of the line is $1$ and the line passes through the origin, or the slope is $-1$. A line with slope $1$ that passes through the origin occurs if $a = d \neq 0$ and $b = 0$. There are $10$ possible values of $a$. The slope is $-1$ if $d = -a \neq 0$. There are $10$ possible values of $a$, and $11$ possible values of $b$, which gives $110$ possibilities. This gives a total of $120$ possibilities in Case 1.

CASE 2. Suppose that $c \neq 0$. The quantity $y = \frac{ax+b}{cx+d}$ tends to $\frac{a}{c}$ as $x \to \pm\infty$. This is a horizontal asymptote. The degree of the polynomial in the numerator can be reduced by subtracting this asymptotic value:
$$y - \frac{a}{c} = \frac{ax + b}{cx + d} - \frac{a}{c} = \frac{acx + bc - a(cx + d)}{c(cx + d)} = \frac{bc - ad}{c(cx + d)} = \frac{\Delta}{cx + d}$$
where
$$\Delta = bc - ad.$$
If $\Delta = 0$, then the graph of $y$ consists of the points $(x, \frac{a}{c})$, except that the point $(-\frac{d}{c}, \frac{a}{c})$ is missing because the formula reads $\frac{1}{0}$ when $x = -\frac{d}{c}$. In any case, this set of points is not symmetric with respect to the line $y = x$. From now on, assume that $\Delta \neq 0$. Multiply both sides by $cx + d$ to find that
$$(cx + d)(y - \frac{a}{c}) = bc - ad.$$
Note that $cx + d \neq 0$ because $\Delta \neq 0$. Thus, if it were desired, both sides of the new equation could be divided by $cx + d$ to recover the former equation. Hence the set of pairs $(x, y)$ that satisfy the former equation is exactly the same as the set of $(x, y)$ that satisfy the new equation, which determines a rectangular hyperbola. This hyperbola is symmetric with respect to the line $y = x$ if and only if its center $(\frac{a}{c}, -\frac{d}{c})$ lies on this line of symmetry. That is, $d = -a$. Thus $\Delta \neq 0$ if and only if $bc + a^2 \neq 0$.

If $a = 0$, then $b$ and $c$ are free to take on any nonzero values, which gives $10 \cdot 10 = 100$ quadruples $(0, b, c, 0)$. If $a \neq 0$, then because $d = -a$ and $c \neq 0$, there are $10 \cdot 11 \cdot 10 = 1100$ quadruples $(a, b, c, -a)$ to consider. Note that $bc + a^2 = 0$ if and only if either $|a| = |b| = |c|$ and $b$ and $c$ have opposite signs, which is $10 \cdot 2 = 20$ cases, or $|a| = 2$ and $(b, c) = (\pm 4, \mp 1)$ or $(\pm 1, \mp 4)$, which is $2 \cdot 2 \cdot 2 = 8$ additional cases. Thus the total number of possibilities in Case 2 is $100+1100-(20+8) = 1172$.

In all there are $120 + 1172 = 1292$ quadruples satisfying the given conditions.