1. Write the parabola in standard form

   Given parabola: $y^2 = 16(x-3)$

   Compare with standard form: $y^2 = 4a(x-h)$

   So, $4a = 16 \Rightarrow a = 4, \qquad h=3$

   Hence the parabola is a rightward opening parabola with vertex at $(3,0)$.

2. Take a general point on the parabola

   For the standard parabola $y^2=4ax$, a parametric point is $(at^2, 2at)$.

   Since our parabola is shifted by $3$ units to the right, a general point on it is: $\big(3+at^2,\; 2at\big)$

   With $a=4$: $\big(3+4t^2,\; 8t\big)$

3. Equation of tangent at parameter $t$

   For $y^2=4a(x-h)$, tangent at parameter $t$ is: $ty = x-h + at^2$

   Here, $ty = x-3 + 4t^2$

   Rearranging: $x - ty + (4t^2-3)=0$

4. Condition that tangent passes through point $P(x_1,y_1)$

   If this tangent passes through $P(x_1,y_1)$, then: $t y_1 = x_1 - 3 + 4t^2$

   or $4t^2 - y_1 t + (x_1-3)=0$

   This quadratic in $t$ gives the parameters of the two tangents drawn from $P$.

   Let the two roots be $t_1$ and $t_2$.

5. Use the condition that the tangents are perpendicular

   For tangent $ty = x-3+4t^2$ we can write $y = \frac{1}{t}x + \frac{4t^2-3}{t}$

   So slope of tangent is: $m = \frac{1}{t}$

   Therefore slopes of the two tangents are: $m_1 = \frac{1}{t_1}, \qquad m_2 = \frac{1}{t_2}$

   For perpendicular tangents: $m_1 m_2 = -1$

   Hence, $\frac{1}{t_1 t_2} = -1 \Rightarrow t_1 t_2 = -1$

6. Relate product of roots to point $P$

   From the quadratic $4t^2 - y_1 t + (x_1-3)=0$

   product of roots is: $t_1 t_2 = \frac{x_1-3}{4}$

   Since $t_1 t_2=-1$, $\frac{x_1-3}{4}=-1$

   $x_1 - 3 = -4$

   $x_1 = -1$

7. Locus of $P$

   Therefore the locus is: $x = -1$ or $x + 1 = 0$

8. Match with options

   This is Option B.

Comparison with stored answer:

Stored correct answer: B

Derived answer: B

So the derived answer agrees with the stored correct answer.