Let the two solid dissociations be of the standard type:

$A(s) \rightleftharpoons B(s) + C(g)$ with equilibrium pressure $x$ atm, and

$P(s) \rightleftharpoons Q(s) + R(g)$ with equilibrium pressure $y$ atm.

Since only solids and gases are involved, the equilibrium constants depend only on gaseous pressures.

1. Meaning of the given pressures $x$ and $y$

For the first dissociation:

$A(s) \rightleftharpoons B(s) + C(g)$

If its equilibrium pressure is $x$ atm, then

$K_{p1} = p_C = x$

Similarly, for the second dissociation:

$P(s) \rightleftharpoons Q(s) + R(g)$

$K_{p2} = p_R = y$

2. When both solids dissociate simultaneously

When both are present together, each solid establishes its own equilibrium independently because solids have unit activity.

Thus at equilibrium:

$p_C = x, \qquad p_R = y$

Hence total pressure should be

$P_{\text{total}} = p_C + p_R = x + y$

3. Checking options

• A: $(x+y)$ atm ✅
• B: $\sqrt{x+y}$ atm ❌
• C: $2\sqrt{x+y}$ atm ❌
• D: $x^2+y^2$ atm ❌

So the correct answer is:

$\boxed{x+y\text{ atm}}$

4. Comparison with stored answer

Stored correct answer = C

My derived answer = A

These do not match.

The stored answer would be correct only for a different type of dissociation, such as when each solid produces two moles of gaseous products and the given $x,y$ were related differently to $K_p$. But for the usual interpretation of independent solid dissociations giving one gaseous species each, total pressure is simply the sum of their equilibrium partial pressures.