relationship between equivalent ways of computing RSA private key
$D={k \phi(N)+1 \over E}$ is equivalent to $ED = k \phi(N) + 1$

$ED \equiv 1 \pmod {\phi(N)}$ means "there's some integer $k$ s.t. $ED = 1 + k \phi(N)$"

Hence, they're equivalent.

BTW: the actual necessary and sufficient condition on E, D actually is $ED \equiv 1 \pmod {\text{lcm}(P-1, Q-1)}$ (for RSA modulii that consists of two primes); the relation you're using is sufficient, but not necessary.