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Set $K=\mathbb{Q}(\sqrt{pq})$ and $E=\mathbb{Q}(\sqrt{p},\sqrt{q})$. The distinct primes give independent square classes, so $[E:\mathbb{Q}]=4$ and $[E:K]=2$. In particular $K(\sqrt{p})=K(\sqrt{q})=E$: these are the same quadratic extension of $K$. We prove it is unramified by an explicit field discriminant calculation.

Since $p,q\equiv 1(\mathrm{mod}4)$, put $a=(1+\sqrt{p})/2$, $b=(1+\sqrt{q})/2$. Both are algebraic integers. The ring $R=\mathbb{Z}[a,b]\subseteq {\mathcal{O}}_E$ has basis $1,a,b,ab$. Up to the ordering of the basis, its trace pairing matrix is the tensor product of the quadratic trace matrices, because the four embeddings of $E$ choose the two radical signs independently. The two quadratic matrices areThusOn the other hand, $pq$ is squarefree and congruent to one modulo four, so $d_K=pq$. Let $j=[{\mathcal{O}}_E:R]$. The discriminant-index formula for an integral lattice and the relative discriminant tower formula giveThe ideal norm on the right is a positive integer, while $j$ is a positive integer. Hence $N({\mathfrak{d}}_{E/K})=1/j^2$ forces $j=1$ and ${\mathfrak{d}}_{E/K}={\mathcal{O}}_K$. Its prime divisors are exactly the ramified finite primes, so $E/K$ is unramified at every finite prime, including primes above $2,p,q$. This also proves $d_E=p^2{q}^2$, agreeing with the discriminant of a biquadratic field formula.

Every embedding of $E$ is real, so the real places of $K$ split as well. The extension is quadratic, hence abelian. The ordinary Hilbert class field characterization therefore impliesFinally, the Artin reciprocity map identifies $\mathrm{Cl}(K)$ with $\mathrm{Gal}(H_K/K)$. Restriction onto $\mathrm{Gal}(E/K)\simeq C_2$ is surjective, so the ideal class group has a quotient of order two. Therefore $\boxed{|\mathrm{Cl}(K)|\text{ is even}}$.