Since similar triangles give the same value of $K P / R^{3}$, we can fix $R=1$ and maximize $K P$ over all triangles inscribed in the unit circle. Fix points $A$ and $B$ on the unit circle. The locus of points $C$ with a given perimeter $P$ is an ellipse that meets the circle in at most four points. The area $K$ is maximized (for a fixed $P$) when $C$ is chosen on the perpendicular bisector of $A B$, so we get a maximum value for $K P$ if $C$ is where the perpendicular bisector of $A B$ meets the circle. Thus the maximum value of $K P$ for a given $A B$ occurs when $A B C$ is an isosceles triangle. Repeating this argument with $B C$ fixed, we have that the maximum occurs when $A B C$ is an equilateral triangle.

Consider an equilateral triangle with side length $a$. It has $P=3 a$. It has height equal to $a \sqrt{3} / 2$ giving $K=a^{2} \sqrt{3} / 4$. From the extended law of sines, $2 R=a / \sin (60)$ giving $R=a / \sqrt{3}$. Therefore the maximum value we seek is

$$K P / R^{3}=\left(\frac{a^{2} \sqrt{3}}{4}\right)(3 a)\left(\frac{\sqrt{3}}{a}\right)^{3}=\frac{27}{4} .$$