Step-by-step Solution:

1. Analyze the given equation of the circle. The equation of the given circle is $S: {x^2} + {y^2} - 6x - 4y - 11 = 0$. To find the center and radius, we compare it with the general equation of a circle, ${x^2} + {y^2} + 2gx + 2fy + c = 0$. We have: $2g = -6 \implies g = -3$ $2f = -4 \implies f = -2$ $c = -11$ The center of the circle, $C$, is given by $(-g, -f)$. So, $C = (3, 2)$.

2. Understand the geometry of the problem. Let $P(1, 8)$ be the external point from which tangents are drawn to the circle. Let the tangents from $P$ touch the circle at points $A$ and $B$. The line segments $CA$ and $CB$ are the radii of the circle at the points of contact. We know that the radius to the point of tangency is perpendicular to the tangent line at that point. Therefore, $CA \perp PA$ and $CB \perp PB$. This means $\angle CAP = 90^\circ$ and $\angle CBP = 90^\circ$.

3. Identify the properties of the quadrilateral PACB. Consider the quadrilateral formed by the points $P, A, C,$ and $B$. The sum of the opposite angles $\angle CAP$ and $\angle CBP$ is: $\angle CAP + \angle CBP = 90^\circ + 90^\circ = 180^\circ$. A quadrilateral is cyclic if the sum of a pair of opposite angles is $180^\circ$. Thus, the quadrilateral $PACB$ is cyclic.

4. Determine the circumcircle of triangle PAB. The circumcircle of the triangle $PAB$ is the circle that passes through the vertices $P, A,$ and $B$. Since the points $P, A, C,$ and $B$ lie on the same circle (the circumcircle of quadrilateral $PACB$), the circumcircle of triangle $PAB$ is the same as the circumcircle of quadrilateral $PACB$.

5. Find the equation of the circumcircle. In the cyclic quadrilateral $PACB$, the angles $\angle PAC$ and $\angle PBC$ are $90^\circ$. These are angles in a semicircle. This implies that the line segment $PC$ is the diameter of the circumcircle of $PACB$. The coordinates of the endpoints of the diameter are $P(1, 8)$ and $C(3, 2)$.

6. Use the diameter form of the equation of a circle. The equation of a circle with the endpoints of a diameter at $(x_1, y_1)$ and $(x_2, y_2)$ is given by: $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$ Substituting the coordinates of $P(1, 8)$ and $C(3, 2)$: $(x - 1)(x - 3) + (y - 8)(y - 2) = 0$ Expanding the terms: $(x^2 - 3x - x + 3) + (y^2 - 2y - 8y + 16) = 0$ $x^2 - 4x + 3 + y^2 - 10y + 16 = 0$ Combining like terms, we get the final equation: ${x^2} + {y^2} - 4x - 10y + 19 = 0$

7. Compare with the given options. The derived equation is ${x^2} + {y^2} - 4x - 10y + 19 = 0$, which matches option B.

Conclusion:

The equation of the circumcircle of the triangle PAB is ${x^2} + {y^2} - 4x - 10y + 19 = 0$.