The work done ($W$) in stretching a spring is stored as elastic potential energy, given by the formula $W = \frac{1}{2}kx^2$, where $k$ is the spring constant and $x$ is the displacement .

1. Case (a): Same displacement ($x_P = x_Q = x$) Using $W = \frac{1}{2}kx^2$, since $x$ is constant, $W \propto k$. Given $k_P > k_Q$, it follows that $W_P > W_Q$.

2. Case (b): Same force ($F_P = F_Q = F$) We know $F = kx$, so $x = F/k$ . Substituting this into the work formula: $W = \frac{1}{2}k(F/k)^2 = \frac{F^2}{2k}$. Here, since $F$ is constant, $W \propto 1/k$. Given $k_P > k_Q$, it follows that $W_P < W_Q$.

Thus, the correct relationships are $W_P > W_Q$ for case (a) and $W_P < W_Q$ for case (b).