In a mixed oxide, the oxide ions form a cubic close-packed (ccp)...

In a ccp lattice of oxide ions, the number of $\text{O}^{2-}$ ions per unit cell is $4$.

The number of tetrahedral voids is $2 \times 4 = 8$.
Number of $\text{P}$ atoms = $\dfrac{1}{8} \times 8 = 1$.

The number of octahedral voids is $4$.
Number of $\text{Q}$ atoms = $\dfrac{1}{2} \times 4 = 2$.

The empirical formula of the mixed oxide is $\text{PQ}_{2}\text{O}_{4}$.

For the crystal to be electrically neutral, the sum of the oxidation states must be zero.
Let the oxidation state of $\text{P}$ be $x$ and that of $\text{Q}$ be $y$.
$1(x) + 2(y) + 4(-2) = 0$
$x + 2y = 8$

Checking the given options:
For $+3$ and $+2$: $3 + 2(2) = 7 \neq 8$
For $+2$ and $+3$: $2 + 2(3) = 8$ (Satisfies the condition)
For $+4$ and $+1$: $4 + 2(1) = 6 \neq 8$
For $+1$ and $+3$: $1 + 2(3) = 7 \neq 8$

Thus, the oxidation states of $\text{P}$ and $\text{Q}$ are $+2$ and $+3$ respectively.

Answer: $+2$ and $+3$