Here is the proof that does not use the Honda-Tate theory.

Notice that the quadratic field $K=Q(\alpha)$ may be viewed as the subfield of the endomorphism algebra $End^0(E)=End(E)\otimes Q$ with the same identity element. Here $End(E)$ is the algebra of ALL endomorphisms of $E$ over an algebraic closure of $F_q$.

1. Suppose that $E$ is ordinary and $T_p(E)$ is its physical $p$-adic Tate module, which is a free $Z_p$-module of rank 1. Let us consider the corresponding $Q_p$-vector space $V_p(E)=T_p(E)\otimes_{Z_p}Q_p$, which is a one-dimensional vector space over the field $Q_p$ of $p$-adic numbers.

By functoriality, there is the natural $Q_p$-algebra homomomorphism

$$K_p=K\otimes_Q Q_p \to End_{Q_p}(V_p(E))=Q_p,$$ which sends $1$ to $1$ and therefore is not zero. Since $K_p$ has $Q_p$-dimension $2>1$, this homomorphism is not injective and therefore $K_p$ is NOT a field, i.e., $p$ splits in $K=Q(\alpha)$.

2. Suppose that $E$ is supersingular. Then $$End(E)\otimes Q_p=End^0(E)\otimes_Q Q_p$$ is a division algebra over $Q_p$ of dimension 4 that contains $K_p=K\otimes_Q Q_p$ as a $Q_p$-subalgebra. Since $End^0(E)\otimes_Q Q_p$ has no zero divisors, $K_p$ also has no zero divisors, i.e., $p$ does NOT split in $K=Q(\alpha)$.

Here is a proof that does not use the Honda-Tate theory.

Notice that the quadratic field $K=Q(\alpha)$ may be viewed as the subfield of the endomorphism algebra $End^0(E)=End(E)\otimes Q$ with the same identity element. Here $End(E)$ is the algebra of ALL endomorphisms of $E$ over an algebraic closure of $F_q$.

1. Suppose that $E$ is ordinary and $T_p(E)$ is its physical $p$-adic Tate module, which is a free $Z_p$-module of rank 1. Let us consider the corresponding $Q_p$-vector space $V_p(E)=T_p(E)\otimes_{Z_p}Q_p$, which is a one-dimensional vector space over the field $Q_p$ of $p$-adic numbers.

By functoriality, there is the natural $Q_p$-algebra homomomorphism

$$K_p=K\otimes_Q Q_p \to End_{Q_p}(V_p(E))=Q_p,$$ which sends $1$ to $1$ and therefore is not zero. Since $K_p$ has $Q_p$-dimension $2>1$, this homomorphism is not injective and therefore $K_p$ is NOT a field, i.e., $p$ splits in $K=Q(\alpha)$.

2. Suppose that $E$ is supersingular. Then $$End(E)\otimes Q_p=End^0(E)\otimes_Q Q_p$$ is a division algebra over $Q_p$ of dimension 4 that contains $K_p=K\otimes_Q Q_p$ as a $Q_p$-subalgebra. Since $End^0(E)\otimes_Q Q_p$ has no zero divisors, $K_p$ also has no zero divisors, i.e., $p$ does NOT split in $K=Q(\alpha)$.