Step 1: Read what is given.
The circle is $x^2+y^2=64$. Comparing with $x^2+y^2=r^2$, we get $r^2=64$, so the radius is $r=8$. The point $P$ is written as an angle, $P\left(\frac{2\pi}{3}\right)$, which is the parametric way of naming a point on the circle.
Step 2: Why a point can be named by an angle.
Every point on $x^2+y^2=r^2$ can be written as $(r\cos\theta,\ r\sin\theta)$. This always satisfies the circle because $r^2\cos^2\theta+r^2\sin^2\theta=r^2$. So the angle alone fixes the point.
Step 3: Find the actual point.
Here $\theta=\frac{2\pi}{3}=120^\circ$. So \[ x_1=8\cos 120^\circ=8\left(-\tfrac{1}{2}\right)=-4,\qquad y_1=8\sin 120^\circ=8\left(\tfrac{\sqrt{3}}{2}\right)=4\sqrt{3}. \] The point of contact is $(-4,\ 4\sqrt{3})$.
Step 4: Recall the tangent rule.
For a circle $x^2+y^2=r^2$, the tangent at a point $(x_1,y_1)$ on it has a neat ready formula: $xx_1+yy_1=r^2$. We just plug in our point and $r^2$.
Step 5: Put the numbers in.
\[ x(-4)+y(4\sqrt{3})=64 \] \[ -4x+4\sqrt{3}\,y=64. \]
Step 6: Make it simple.
Divide every term by $-4$ to get small numbers: \[ x-\sqrt{3}\,y=-16 \] which we write as $x-\sqrt{3}\,y+16=0$.
Step 7: Pick the option.
This is exactly option (4).
\[ \boxed{x-\sqrt{3}\,y+16=0} \]