Equilibrium constant
The position of equilibrium depends on temperature and pressure, as Nick Burns describes in his answer. The equilibrium constant depends on temperature only. The equilibrium constant is not a measure of the position of equilibrium, but rather which positions (representing different compositions) along the the reaction coordinate surface are equilibrium positions. Changing the temperature changes the shape of the reaction coordinate surface. Changing the pressure just moves you to a different point on the surface. 

Example: consider a hypothetical reaction in which two molecules of gas react to form a third. 

$$\ce{A(g) + B(g)&lt;=&gt;C(g)}$$

The reaction quotient $Q$ is simply the ratio of the product of the partial pressure(s) of the product(s) to the product of the partial pressure(s) of the reactant(s). $Q$ can have any value representing different points on the reaction coordinate surface. 

$$Q=\frac{P_C}{P_A P_B}$$

The equilibrium constant represents special values of Q in which the reaction coordinate surface has a free energy minimum. For the sake of our example, let's make the equilibrium constant an easy value to work with, say $K=2$. 

$$K=Q_{eq}=\frac{P_{C,eq}}{P_{A,eq} P_{B,eq}}=2$$

Since we are talking about gases, we also have the constraint the sum of the partial pressures of the reagents must equal the total pressure of the system. Again, for ease, let us set this value to an easy value to work with, say $P_T = 7 \text{ kPa}$.

$$\frac{P_{C}}{P_{A} P_{B}}=2$$
$$P_A + P_B + P_C = 7$$

Now we have two equations of three variables, which means there are an infinite number of possible values of the partial pressures of the reactions components that fit these two conditions. One is $P_A = 1$, $P_B =2$, and $P_C =4$. Another equally valid combo is $P_A = 2$, $P_B=1$, and $P_C=4$. The position of the equilibrium in this case is better given by a ratio of the sum of the partial pressure(s) of the product(s) to the sum of the partial pressure(s) of the reactant(s):

$$\frac{P_C}{P_A + P_B}=\frac{4}{3}$$

If we change the total pressure to, say $P_T = 4$, we change the allowable combinations of $P_A$, $P_B$, and $P_C$ that can get us to the equilibrium condition of $Q=2$. One allowable combination (of an infinite number) is $P_A=1$, $P_B=1$, and $P_C=2$. The position of the equilibrium is now:

$$\frac{P_C}{P_A + P_B}=\frac{2}{2}=1$$

Le Chatelier's principle suggests that When pressure decreases, the equilibrium shifts toward the side of the reaction with more equivalents of gas. 

If we consider changing the situation from $P_A = 1$, $P_B =2$, $P_C =4$, and $P_T=7$ to a situation with $P_T=2$, we can do some math to figure out new values of $P_A$, $P_B$, and $P_C$ that fit this new situation. First, we set the partial pressures of the compoents to $\frac{2}{7}$ of their original value, and calculate Q.
$$P_A =1\times \frac{2}{7}=\frac{2}{7}$$
$$P_B =2\times \frac{2}{7}=\frac{4}{7}$$
$$P_A =4\times \frac{2}{7}=\frac{8}{7}$$
$$Q=\frac{P_C}{P_A P_B}=\frac{\frac{8}{7}}{\frac{2}{7} \times \frac{4}{7}}=\frac{\frac{8}{7}}{\frac{8}{49}}=49$$

Since $C&gt;K$ the equilibrium will shift to decrease $Q$. This can happen by decreasing $P_C$ and increasing $P_A$ and $P_B$. Since there is an reaction equation relating these three components in a 1:1:1 stoichiometric ratio, the change must have the same value vut in different directions:

$$K=2= \frac{\frac{8}{7}-x}{\left(\frac{2}{7}+x \right)\left( \frac{4}{7}+x\right)}$$

Rather than goof up the algebra, I let WolframAlpha do it. The solution for $x$ is:

$$x=\frac{1}{128}\left(-19 \pm \sqrt{681} \right)\approx \{+0.2534..., -1.6106...\}$$

The first answer for $x$ is nonsense, since it would put one partial pressure less than $0$. If we change the partial pressures appropriately, we get new values of 

$$P_A = \frac{2}{7}+\frac{1}{128}\left(-19 + \sqrt{681} \right)=0.5391...$$
$$P_B = \frac{4}{7}+\frac{1}{128}\left(-19 + \sqrt{681} \right)=0.8248...$$
$$P_C = \frac{8}{7}-\frac{1}{128}\left(-19 + \sqrt{681} \right)=0.8894...$$

The position of equilibrium is

$$\frac{0.8894...}{0.5391...+0.8248...}\approx 0.65$$

which is less than the value of $1.33$ from when $P_T=7$ and different from the value of $1$, when the partial pressures were different (but still met the criteria than $P_T=2$ and $Q=2$).

Changing the temperature changes the values of $Q$ which define the equilibrium state:

$$K=e^{-\Delta G^\circ/RT}$$