Edit:
In my original answer, I ignored the condition that the ace and king are in the same suit. That is partly due to the awkward language, which technically can support my reading. But there is no reason for the question to mention suits at all if the problem was about just having an ace and a king,
so I'm taking the suggestion by commenter AnkitSeth that the correct reading that we are looking for the probability that the hand has, for at least one suit, the ace and king of that suit.
Let $A$ be the set of all deals.
Let $A_i$ be the deals that contain the ace and king of suit $i,$ for $i=1,2,3,4.$
Then the inclusion-exclusion gives the probability:
$$\frac{\binom{4}{1}|A_1|-\binom{4}{2}|A_1\cap A_2|+\binom{4}{3}|A_1\cap A_2\cap A_3|-\binom{4}{4}|A_1\cap A_2\cap A_3\cap A_4|}{|A|}\\=\frac{\binom{4}{1}\binom{50}{11}-\binom{4}{2}\binom{48}{9}+\binom{4}{3}\binom{46}{7}-\binom{4}{4}\binom{44}{5}}{\binom{52}{13}}$$
Original Answer
This answer finds the probability that the hand has one ace and one king.
If $p$ is the probability that you get at least one ace and one king, consider $$1-p = p_{\lnot A}+p_{\lnot K}-p_{\lnot AK}$$ where $p_{\lnot A}$ is that probability that you didn't get any ace, $p_{\lnot K}$ is the probability that you got no king, and $p_{\lnot AK}$ is the probability that you got no ace and no king.
So $$p_{\lnot A} = p_{\lnot K}=\frac{\binom{48}{13}}{\binom{52}{13}}$$ and $$p_{\lnot AK}=\frac{\binom{44}{13}}{\binom{52}{13}}$$
So $$p = \frac{\binom{52}{13} - 2\binom{48}{13} + \binom{44}{13}}{\binom{52}{13}}$$
There are indeed $\binom{52}{13}$ different 13-card hands and this will indeed be the size of our sample space and thus our denominator when we finish our calculations.
For the numerator, we need to pause for a moment and understand what the problem is actually asking, since this appears to be where you got stuck.
We are asked to find the probability that in our hand of thirteen cards, there is at least one suit for which we have all three face cards. For example $(A\spadesuit,2\spadesuit,3\spadesuit,\dots,10\spadesuit,J\spadesuit,Q\spadesuit,K\spadesuit)$ has all three of the face cards for spades. Similarly if all those cards happened to be hearts instead it would also count since we would have all of the face cards for hearts. Similarly still, a hand like $(J\spadesuit,Q\spadesuit,K\spadesuit,J\heartsuit,Q\heartsuit,K\heartsuit,J\diamondsuit,Q\diamondsuit,K\diamondsuit,\dots)$ would count since we have all of the face cards from spades (we also happen to have all of the face cards from hearts and diamonds too).
Let $\spadesuit$ represent the event that we have have all of the face cards from spades. Similarly, let $\diamondsuit, \heartsuit, \clubsuit$ represent the event that we have all of the face cards from diamonds, hearts, and clubs respectively.
You are asked to find $Pr(\spadesuit\cup\diamondsuit\cup \clubsuit\cup \heartsuit)$
To do this, let us apply inclusion exclusion. We expand the above as:
$Pr(\spadesuit\cup \diamondsuit\cup\clubsuit\cup\heartsuit) = Pr(\spadesuit)+Pr(\diamondsuit)+\dots-Pr(\spadesuit\cap \diamondsuit)-Pr(\spadesuit\cap \clubsuit)-\dots+Pr(\spadesuit\cap \diamondsuit\cap \clubsuit)+\dots-Pr(\spadesuit\cap\diamondsuit\cap \clubsuit\cap \heartsuit)$
Now, let us calculate each individual term in the expansion.
The calculation you did before is relevant. Indeed, we calculate $Pr(\spadesuit)=\dfrac{\binom{3}{3}\binom{49}{10}}{\binom{52}{13}}$. This is again merely the probability that we have all of the face cards from the spades and is not the final probability that we were tasked with calculating.
We continue and calculate more terms:
For example $Pr(\spadesuit\cap \diamondsuit)=\dfrac{\binom{6}{6}\binom{46}{7}}{\binom{52}{13}}$
We then notice what symmetry there is in the terms and can simplify some. Finally, we write the final expression for our final answer (and get an exact number only if actually requested or required, usually opting to leave the answer in terms of binomial coefficients without additional simplification).
Best Answer
It is as easy as PIE (the Principle of Inclusion and Exclusion).$$\mathsf P(A\cup B)=\mathsf P(A)+\mathsf P(B)-\mathsf P(A\cap B)$$
The sample space contains $\binom{13}5$ equally probable ways to select five cards from the small deck, without replacement or bias.
There are $\binom {11}3$ ways to select the Jack, the Queen, and three other cards from the small deck.
Likewise there are $\binom{11}3$ ways to select the 10, the Jack, and three other cards from the small deck.
But wait! We do not wish to over-count, and there are $\binom {10}2$ ways to select the 10, the Jack, and the Queen from the small deck.
Thus the probability for our favoured event is:$$\dfrac{2\binom {11}3-\binom {10}2}{\binom{13}5}$$
Remark: Using a small deck makes this easy. We do not have to worry about the possibility of pairs or such; there is only one card for each kind in the deck.