Is a knot $K$ and it's mirror image $^*K$ considered the same knot in terms of tabulating prime knots? If so, why? I'm just wanting to confirm whether this is the case and why? Is it purely to do with the sheer number of knot projections that would have to be dealt with?
 A: A knot and its mirror image are not always the same, but when it comes to knot tables, most people do not bother drawing both.  I believe this is just for space and the fact that we can pretty much visualize what the mirror image looks like. But there is a way to tell. 
Most tables will then provide the Jones Polynomial of the knot, which can distinguish mirror images.  It is usually given by some sequence of numbers like: $\{-4\}(-1, 1, 0, 1)$ which means a polynomial with $-4$ exponent as the leading term and $-1$ as its coefficient. So, the actual polynomial is $-t^{-4}+t^{-3}+t^{-1}$, which is the trefoil. The impressive fact that if you substitute $t^{-1}$ in for $t$, it will be the polynomial of the mirror image, is why the Jones polynomial is so powerful. If the two polynomials you get this way are not equal, the mirror images are not equivalent. Notice, they are not equivalent for the trefoil.
But for the figure eight knot, $\{-2\}(1,-1,1,-1,1)$ is symmetric, so by the Jones polynomial, it may be equivalent to its mirror image, as pointed out by Grumpy Parsnip, and in this case, it is.
