If $$5x + 9 = 0$$ is the directrix of the hyperbola $$16x^2 - 9y^2 = 144$$, then its corresponding focus is:

We are given the hyperbola $$16x^2-9y^2=144$$. To bring it into the standard form, we divide every term by $$144$$:

$$\frac{16x^2}{144}-\frac{9y^2}{144}=1 \;\Longrightarrow\; \frac{x^2}{9}-\frac{y^2}{16}=1.$$

Now the equation is of the form $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$ with

$$a^2=9,\; b^2=16 \;\Longrightarrow\; a=3,\; b=4.$$

For a rectangular hyperbola oriented along the $$x$$-axis, the distance of each focus from the centre satisfies the relation

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

Substituting $$a^2=9$$ and $$b^2=16$$ we obtain

$$c^2=9+16=25 \;\Longrightarrow\; c=5.$$

The eccentricity $$e$$ is defined by the formula

$$e=\frac{c}{a}.$$

Hence

$$e=\frac{5}{3}.$$

For the hyperbola $$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1,$$ the directrices are given by the pair of vertical lines

$$x=\pm\frac{a}{e}.$$

Substituting $$a=3$$ and $$e=\dfrac{5}{3},$$ we get

$$\frac{a}{e}=\frac{3}{\frac{5}{3}}=\frac{9}{5}.$$

So the two directrices are

$$x=\frac{9}{5}\quad\text{and}\quad x=-\frac{9}{5}.$$

The problem states that the directrix is $$5x+9=0,$$ which can be rewritten as

$$x=-\frac{9}{5}.$$

This matches the negative directrix $$x=-\dfrac{9}{5}.$$ The corresponding focus lies on the same (negative) side of the centre along the transverse axis. The foci for this hyperbola are $$(\pm c,0)=(\pm5,0).$$ Therefore, the required focus is

$$(-5,0).$$

Hence, the correct answer is Option A.