If the co-ordinates of two points $$A$$ and $$B$$ are $$\left(\sqrt{7}, 0\right)$$ and $$\left(-\sqrt{7}, 0\right)$$ respectively and $$P$$ is any point on the conic, $$9x^2 + 16y^2 = 144$$, then $$PA + PB$$ is equal to:

We are given that point $$P(x,y)$$ lies on the conic $$9x^2 + 16y^2 = 144$$.

First we rewrite the conic in standard ellipse form. Dividing every term by $$144$$, we obtain

$$\dfrac{9x^2}{144} + \dfrac{16y^2}{144} = 1.$$

Simplifying the fractions,

$$\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1.$$

We recognise $$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$$ as the standard equation of an ellipse centred at the origin with semi-major axis $$a$$ and semi-minor axis $$b$$. Comparing, we have

$$a^2 = 16 \quad\Rightarrow\quad a = 4, \qquad b^2 = 9 \quad\Rightarrow\quad b = 3.$$

For any ellipse, the distance from the centre to each focus (denoted $$c$$) satisfies the relation

$$c^2 = a^2 - b^2.$$

Substituting our values $$a^2 = 16$$ and $$b^2 = 9$$, we get

$$c^2 = 16 - 9 = 7 \;\;\Longrightarrow\;\; c = \sqrt{7}.$$

Hence the foci are located at $$(c,0)$$ and $$(-c,0),$$ namely $$(\sqrt{7},0)$$ and $$(-\sqrt{7},0).$$ These are exactly the points $$A$$ and $$B$$ given in the question.

Now we recall the fundamental property of an ellipse:

The sum of the distances from any point on an ellipse to its two foci is constant and equals $$2a$$.

Here, $$a = 4,$$ so

$$PA + PB = 2a = 2 \times 4 = 8.$$

Therefore the value of $$PA + PB$$ is a constant $$8$$ for every point $$P$$ on the given ellipse.

Hence, the correct answer is Option B.