a. Suppose $\mathbb{Q}(\sqrt p)$ is isomorphic to $\mathbb{Q}(\sqrt q)$ as fields. Then $\mathbb{Q}(\sqrt q)$ has an element $\alpha$ such that $\alpha^2 = p$. Let $\alpha = a + b\sqrt q$, where $a, b \in \mathbb{Q}$. We denote the conjugate of $\alpha$ by $\alpha'$. Since $\alpha + \alpha' = 0$, $a = 0$. Since $\alpha^2 = p$, $p = b^2 q$. Hence $p = q$. This is a contradiction. Hence $a$. is not true.
b. Since the both fields have dimension 2 as vector spaces over $\mathbb{Q}$, $b.$ is true.
c. As we see in the above, $\sqrt p$ is not contained in $\mathbb{Q}(\sqrt q)$. Hence $[\mathbb{Q}(\sqrt p,\sqrt q) : \mathbb{Q}(\sqrt q)] = 2$. Hence $[\mathbb{Q}(\sqrt p,\sqrt q) : \mathbb{Q}] = 4$. Thus $c.$ is true.
d. Let $K = \mathbb{Q}(\sqrt p,\sqrt q)$. Clearly $K/\mathbb{Q}$ is Galois. Let $\sigma \in Gal(K/\mathbb{Q})$. Then $\sigma(\sqrt p) = \sqrt p$ or $-\sqrt p$, $\sigma(\sqrt q) = \sqrt q$ or $-\sqrt q$. Since $|Gal(K/\mathbb{Q})| = 4$ by $c.$, $Gal(K/\mathbb{Q}) = \{1, \sigma_1, \sigma_2. \sigma_3\}$, where
$\sigma_1(\sqrt p) = \sqrt p,\ \ \ \sigma_1(\sqrt q) = -\sqrt q$
$\sigma_2(\sqrt p) = -\sqrt p,\ \sigma_2(\sqrt q) = \sqrt q$
$\sigma_3(\sqrt p) = -\sqrt p,\ \ \ \sigma_3(\sqrt q) = -\sqrt q$
Let $\alpha = \sqrt p + \sqrt q$.
Clearly $\alpha, \sigma_1(\alpha) = \sqrt p - \sqrt q, \sigma_2(\alpha) = -\sqrt p + \sqrt q, \sigma_3(\alpha) = -\sqrt p - \sqrt q$ are distinct. Hence $K = \mathbb{Q}(\alpha)$. Thus $d.$ is true.