$N{H_3}COON{H_2}(s)\underset {} \leftrightarrows 2N{H_3}(g) + C{O_2}(g)$.
If equilibrium pressure is 3 atm for the given reaction, then Kp will be:
A.4
B.27
C.4/27
D.1/27

Hint: ${K_p}$ is the equilibrium constant in a gaseous form which is equal to the ratio of the multiple of the partial pressure of products to the multiple of the partial pressure of the reactants, each raised to the power their stoichiometric coefficient. In this numerical the formula used is given below:
${K_p} = \dfrac{{{p_X} \times {p_Y}}}{{{p_A} \times {p_B}}}$
Where A, B are the reactants, and X, Y are the products.
$N{H_3}COON{H_2}(s)\underset {} \leftrightarrows 2N{H_3}(g) + C{O_2}(g)$

Complete step by step answer:
As we know at equilibrium, if volume and temperature are constant then the number of the moles of a gas is directly proportional to its partial pressure. Therefore, the partial pressure of 2 moles of ammonia is equal to 2p. Similarly, the partial pressure of one mole of carbon dioxide gas is equal to p. Hence the total pressure = 2p+p = 3p
Given, total pressure = 3
On comparing the above two equation we get,
3p=3
Hence p=1
So by using the formula, ${k_p} = \dfrac{{p_{N{H_3}}^2 \times {p_{C{O_2}}}}}{{{p_{N{H_3}COON{H_2}}}}}$
$ \Rightarrow {k_p} = \dfrac{{{{(2p)}^2} \times (p)}}{{(p)}}$
$ \Rightarrow {k_p} = 4{p^2}$
Putting the value of p that is equal to 1, we get:
$
  {k_p} = 4 \times {(1)^2} \\
  {k_p} = 4 \\
$Hence the correct answer is (A).

Note:
One more type of equilibrium constant exists i.e. Kc means equilibrium constant in terms of molar concentration.
$k_c$ and $k_p$ are related to each other as:
${k_p} = {k_c}.{(RT)^{\Delta n}}$
Where R is the gas constant and T is the temperature and $\Delta n$ is the (number of moles of gaseous products- number of the moles of the reactants).