Step-by-step Solution:

1. Identify the Ellipse and the External Point The equation of the ellipse is given by ${{{x^2}} \over 9} + {{{y^2}} \over 4} = 1$. Here, $a^2 = 9$ and $b^2 = 4$. The external point is $P(3, 4)$. Tangents are drawn from $P$ to the ellipse, touching it at points $A$ and $B$.

2. Find the Equation of the Chord of Contact AB The equation of the chord of contact from an external point $(x_1, y_1)$ to the ellipse ${{{x^2}} \over {a^2}} + {{{y^2}} \over {b^2}} = 1$ is given by the formula $T=0$, which is ${{{xx_1}} \over {a^2}} + {{{yy_1}} \over {b^2}} = 1$. For the point $P(3, 4)$, the equation of the chord of contact $AB$ is: ${{x(3)} \over 9} + {{y(4)} \over 4} = 1$ ${x \over 3} + y = 1$ $x + 3y = 3$

3. Find the Coordinates of the Points of Contact A and B To find the coordinates of $A$ and $B$, we solve the equation of the ellipse and the equation of the chord of contact simultaneously. From the chord equation, we have $x = 3 - 3y$. Substituting this into the ellipse equation: ${{(3 - 3y)^2} \over 9} + {{y^2} \over 4} = 1$ ${{9(1 - y)^2} \over 9} + {{y^2} \over 4} = 1$ $(1 - y)^2 + {{y^2} \over 4} = 1$ $1 - 2y + y^2 + {{y^2} \over 4} = 1$ $-2y + {{5y^2} \over 4} = 0$ $y\left( {{5y \over 4} - 2} \right) = 0$ This gives two possible values for $y$: $y = 0$ or ${5y \over 4} = 2 \implies y = {8 \over 5}$.

   Now, we find the corresponding $x$ coordinates using $x = 3 - 3y$:

   • If $y = 0$, then $x = 3 - 3(0) = 3$. So, point $A$ is $(3, 0)$.
   • If $y = {8 \over 5}$, then $x = 3 - 3\left({8 \over 5}\right) = 3 - {{24} \over 5} = {{15 - 24} \over 5} = -{9 \over 5}$. So, point $B$ is $\left(-{9 \over 5}, {8 \over 5}\right)$.

4. Find the Orthocentre of Triangle PAB The vertices of the triangle are $P(3, 4)$, $A(3, 0)$, and $B\left(-{9 \over 5}, {8 \over 5}\right)$. The orthocentre is the point where the altitudes of the triangle intersect.

   Altitude from B to PA: The side $PA$ connects points $P(3, 4)$ and $A(3, 0)$. Since the x-coordinates are the same, the line $PA$ is a vertical line with the equation $x = 3$. The altitude from vertex $B$ to the side $PA$ must be a horizontal line. The equation of this altitude is $y = y_B$, which is: $y = {8 \over 5}$

   Altitude from P to AB: The side $AB$ has the equation $x + 3y = 3$. The slope of this line is $m_{AB} = -{1 \over 3}$. The altitude from $P$ to $AB$ will be perpendicular to $AB$. The slope of this altitude, $m_{alt}$, is the negative reciprocal of $m_{AB}$: $m_{alt} = -{1 \over m_{AB}} = -{1 \over -1/3} = 3$ This altitude passes through point $P(3, 4)$. Using the point-slope form $y - y_1 = m(x - x_1)$: $y - 4 = 3(x - 3)$ $y - 4 = 3x - 9$ $3x - y = 5$

5. Calculate the Intersection of the Altitudes To find the orthocentre, we find the intersection of the two altitudes we found: Altitude 1: $y = {8 \over 5}$ Altitude 2: $3x - y = 5$

   Substitute the value of $y$ from the first equation into the second equation: $3x - \left({8 \over 5}\right) = 5$ $3x = 5 + {8 \over 5}$ $3x = {{25 + 8} \over 5} = {{33} \over 5}$ $x = {{33} \over {5 \times 3}} = {{11} \over 5}$

   So, the coordinates of the orthocentre are $\left({{11} \over 5}, {8 \over 5}\right)$.

6. Conclusion The orthocentre of the triangle $PAB$ is $\left({{11} \over 5}, {8 \over 5}\right)$. This matches option C.