Okay, as @JMoravitz commented, there is a much easier method.
However, since you desire to do so using Bayes' Rule:
The deck is partitioned into four parts: (A) the two places in your hand, (B) the three places face up, (C) the next place in the deck, and (D) the forty-six remaining places of the deck.
The cards are partitioned into three categories : (I) the king of hearts and eight of spades, (II) the six remainder of kings and eights, (III) the remaining forty-four cards.
The events are: $E$ the evidence, and $F$ the favored event.
The evidence is that the two places of $A$ contains both from the type-$\sf I$ cards, the three places of $B$ contains none from the type-$\sf II$ cards ... and therefore three from the type-$\sf III$ cards, with the remaining cards are split among the forty-seven places of $C$ and $D$ . This is evaluated using multinomial coefficients as:
$$\begin{align}\mathsf P(E) ~&=~\left.\dbinom{2}{2,0,0}\dbinom{3}{0,0,3}\dbinom{1+46}{0,6,41}\middle/\dbinom{52}{2,6,44}\right.
\end{align}$$
The favoured event is that place $C$ contains 1 from the type-$\sf II$ cards ... with the remaining cards split among the 51 places of A, B, and D.
Well, that should suffice to get you started on finding $\mathsf P(F\mid E)$ using Bayes' Rule.
Best Answer
You don't have to compute $P(B)$. Just deal everybody else out of a $50$ card deck with $11$ hearts. There are $39$ non-hearts. Dealing $18$ cards with no hearts is then a probability of $\frac{39\cdot 38\cdot 37\cdot \dots 22}{50\cdot 49 \cdot 48 \cdot \dots 33}=\frac {39!32!}{50!21!}$