Split up like $52=4+4+44$, i.e. $4$ kings $4$ aces and the other $44$ cards.
In total $N:=\binom{52}2$ different pair-picks are possible and equiprobable.
- $\binom40\binom40\binom{44}2$ of them give $(K,A)=(0,0)$
- $\binom40\binom41\binom{44}1$ of them give $(K,A)=(0,1)$
- $\binom41\binom40\binom{44}1$ of them give $(K,A)=(1,0)$
- $\binom40\binom42\binom{44}0$ of them give $(K,A)=(0,2)$
- $\binom41\binom41\binom{44}0$ of them give $(K,A)=(1,1)$
- $\binom42\binom40\binom{44}0$ of them give $(K,A)=(2,0)$
You can find the probabilities by dividing the outcomes by $N$.
edit (alternative):
Pick two cards one by one.
For $i=1,2$ let $A_{i},K_{i},E_{i}$ denote the events that the $i$-pick
gives ace, king and other card respectively.
$P\left(A_{1}\cap A_{2}\right)=P\left(A_{1}\right)P\left(A_{2}\mid A_{1}\right)=\frac{4}{52}\frac{3}{51}$
$P\left(A_{1}\cap K_{2}\right)=P\left(A_{1}\right)P\left(K_{2}\mid A_{1}\right)=\frac{4}{52}\frac{4}{51}$
$P\left(A_{1}\cap E_{2}\right)=P\left(A_{1}\right)P\left(E_{2}\mid A_{1}\right)=\frac{4}{52}\frac{44}{51}$
$P\left(K_{1}\cap A_{2}\right)=P\left(K_{1}\right)P\left(A_{2}\mid K_{1}\right)=\frac{4}{52}\frac{4}{51}$
$P\left(K_{1}\cap K_{2}\right)=P\left(K_{1}\right)P\left(K_{2}\mid K_{1}\right)=\frac{4}{52}\frac{3}{51}$
$P\left(K_{1}\cap E_{2}\right)=P\left(K_{1}\right)P\left(E_{2}\mid K_{1}\right)=\frac{4}{52}\frac{44}{51}$
$P\left(E_{1}\cap A_{2}\right)=P\left(E_{1}\right)P\left(A_{2}\mid E_{1}\right)=\frac{44}{52}\frac{4}{51}$
$P\left(E_{1}\cap K_{2}\right)=P\left(E_{1}\right)P\left(K_{2}\mid E_{1}\right)=\frac{44}{52}\frac{4}{51}$
$P\left(E_{1}\cap E_{2}\right)=P\left(E_{1}\right)P\left(E_{2}\mid E_{1}\right)=\frac{44}{52}\frac{43}{51}$
As an example note that e.g. $$P(A=1,K=1)=P(A_1\cap K_2)+P(K_1\cap A_2)=\frac{4}{52}\frac{4}{51}+\frac{4}{52}\frac{4}{51}$$