The number of draws until keeping four cards is $T=T_1+T_2+T_3+T_4$ where $T_i$ is the number of draws after obtaining the $(i-1)$th card until obtaining the $i$th card.
The expected number of draws until getting the first card is $E[T_1] = \frac{52}{8}$.
The expected number of draws until getting the next card is $E[T_2] = \frac{51}{7}$.
From here, some casework is necessary. There is a $\frac{3}{7}$ chance that the second kept card is the same rank as the first kept card (call this event $A$), and a $\frac{4}{7}$ chance that this second kept card is the other rank (J if the first kept card was K, and vice versa). In the first case, you are now only pursuing cards of the missing rank; in the second case you are still open to receiving either rank.
So,
$$E[T_3] = E[T_3 \mid A] P(A) + E[T_3 \mid A^c] P(A^c)
= \frac{50}{4} \cdot \frac{3}{7} + \frac{50}{6} \cdot \frac{4}{7}.$$
Finally $E[T_4] = \frac{49}{3}$.