Probability that each queen is drawn before the king of the same suit, given 4 kings and 4 queens in a deck

I am trying to solve the problem from the Hugh Gordon's "Discrete Probability" book (Ch.3, problem 43.c)

Question

We select from a deck of cards the four kings and the four queens. From these eight cards, we draw one card at a time, without replacement, until all eight cards have been drawn. Find the probability that each queen is drawn before the king of each suit.

My attempt of a solution

I realize that there are $8!$ possible arrangements of four kings and four queens. My initial reaction was that if I treat each pair of QK of the respective suit as the distinct unit then I can arrange the 4 pairs in $4!$ ways . The answer at the back of a book is $1/16$, which suggests that my estimate of $p=\frac{4!}{8!}$ does not count additional options, such as the case $QsQcQhQdKcKdKsKh$, where all queens precede all kings, which happens in $4!\cdot 4!$ ways. But there are other cases. What would be the easiest way to group the cases and count them or am I missing a more trivial approach to this?

Update
I was inspired to reconsider the problem by watching the reasoning from AOPS probability of brother and sister sitting together in a row

Let's focus first on the specific suit, say clubs, and let's treat the remaining distinct cards separately. We have $Qc$ and $Kc$ and the remaining $6$ cards. There are 8 slots that correspond to the order of the draw. Now, in any arrangement of the $6$ remaining non-club cards there will be $2$ slots left "waiting" to be filled by $Qc$ and $Kc$. Vividly, for each such arrangement, either the $Qc$ comes before $Kc$, or the vice versa. Thus, the probability that the $Qc$ comes before $Kc$ is $\frac{1}{2}$.
Of course, the same analysis is true regardless of a suit. Furthermore, if we assume the independence of all these arrangements, then the probability that each queen is drawn before the king of each suit, i.e., the probability of all of the 4 events happening together, is the product of the above probabilities, or $p=(\frac{1}{2})^4=1/16$

In half the cases the Q of spades comes before the K of spades and in half the cases it's the other way around.

And in half of the cases where the Q of spades comes before the K spades, we will have the Q clubs come before the K of clubs.

So in one-fourth of the cases we will have the Q of spades before the K of spades and we will have the Q of clubs before the K of clubs. In 3/4 of the cases one or the other will fail to happen. Now we don't give an At's Rass about whether the the order is QsKsQcKc or QsQcKsKc or QsQcKcKs$ or QcQsKsKc or whetever.... all we care about is that we have --Qs--Ks-- (where - is zero or more cards) and --Qc--Kc--- and we don't care about how the spades interact with the clubs.

As for any arrangement of the 6 other cards will have exactly half with the Q of a specific suit before the K of that suit (and the other half will have the K before the Q) we will have $(\frac 12)^4$ of the $8!$ possible arrangements with each the queens before the specific kings.

.....

Or to be more calculating.

There are ${8\choose 2}$ ways of selecting the $2$ spaces for the SPADES. You must place the Queen in the first place and the King in the second.

There are ${6\choose 2}$ ways of selecting the $2$ spaces from the remaining $6$ spaces for the CLUBS. and ${4\choose 2}$ ways of selecting spaces from the $4$ remaining spaces for the diamonds. ANd $2\choose 2$ ways for the HEARTS.

That is ${8\choose 2}{6\choose 2}{4\choose 2}{2\choose 2}$ out of $8!$ ways to do it

$\frac {{8\choose 2}{6\choose 2}{4\choose 2}{2\choose 2}}{8!} =$

$\frac {\frac {8!}{6!2!}\frac {6!}{4!2!}\frac {4!}{2!2!}\frac {2!}{0!2!}}{8!}=$

$\require{cancel)}$

$\frac {\frac {\cancel{8!}}{\cancel{6!}2!}\frac {\cancel{6!}}{\cancel{4!}2!}\frac {\cancel{4!}}{\cancel{2!}2!}\frac {\cancel{2!}}{0!2!}}{\cancel{8!}}=$

$\frac 1{16}$