Calculating partial pressure equilibrium constant Kₚ given initial pressure and equilibrium pressure
The problem is that you are using wrong pressures.
By definition for the reaction at equilibrium partial pressures can be expressed via initial partial pressure $P_1$ and conversion factor $α$

$$
\begin{array}{ccc}
\ce{&amp;A(g) &amp;&lt;=&gt; &amp;B(g) &amp;+ &amp;C(g)}\\
&amp;(1 - α)P_1&amp; &amp; αP_1&amp; &amp; αP_1
\end{array}
$$

equilibrium constant $K_p$ is to be found as

$$K_p = \frac{P(\ce{B})\cdot P(\ce{C})}{P(\ce{A})} = \frac{α^2P_1^2}{(1 - α)P_1} = \frac{α^2P_1}{1 - α}$$

Unknown $α$ can be found by equating total pressure at equilibrium $P_2$ to the sum of partial pressures of all gaseous components at equilibrium:

$$
\begin{align}
P_2 &amp;= P(A) + P(B) + P(C) \\
  &amp;= (1 - α)P_1 + αP_1 + αP_1 \\
  &amp;= (1 + α)P_1 
\end{align}
\quad\implies\quad
α = \frac{P_2}{P_1} - 1 = \frac{\pu{10 atm}}{\pu{6 atm}} - 1 = \frac{2}{3}
$$

Finally, all the values can be plugged into the expression for $K_p$:

$$K_p = \frac{\left(\frac{2}{3}\right)^2\cdot\pu{6 atm}}{\left(1 - \frac{2}{3}\right)} = \pu{8 atm}$$