What is the relationship between the constant of equilibrium (Kc) and the position of equilibrium?
What you are probably referring to is the degree of dissociation $\alpha$ when you ask about the position of equilibrium. At a given temperature the equilibrium constant is just that, constant, but as the pressure changes, say for a gaseous reaction then $\alpha$ changes.An example shows this.

In the gas phase reaction $\ce{A2 &lt;=&gt; 2A}$ with partial pressures $p_{A_2}$ and $p_{A}$ respectively, the  equilibrium constant is $\displaystyle K_p=\frac{p_A^2}{P_{A_2}}$ and is defined at standard conditions of 1 atm.  If $P$ is the total pressure then the two partial pressures are $\displaystyle p_A=\frac{2\alpha}{1+\alpha}P$ and $\displaystyle p_{A_2}=\frac{1-\alpha}{1+\alpha}P$ where and  so after some simplification $\displaystyle K_p=\frac{4\alpha^2}{1-\alpha^2}P$.

Thus, as the equilibrium constant is constant at any total pressure $P$, the degree of dissociation must change as the pressure changes to compensate, thus the 'position of equilibrium' changes. 

[Notes: The amount of A$_2$ reacted is $1-\alpha$ and the amount of A produced is $2\alpha$, thus the mole fractions of each are $(1-\alpha)/(1+\alpha)$ and  $(2\alpha)/(1+\alpha)$ and so the partial pressure are as given above. The total amount needed for the mole fraction is $1-\alpha+2\alpha=1+\alpha$. You can derive related expressions in terms of concentrations.]