Preparation of KHSO4
The pKa2 of sulfuric acid is 1.99 (Table of pka).  This is definitely less acidic than pKa1, but calling this a 'weak' acid may be a distraction.

The relevant question is: does $\ce{KHCO3}$ readily deprotonate $\ce{KHSO4}$?  The pKa1 of carbonic acid is 6.35 (same source), making it a weaker acid than  $\ce{KHSO4}$.  The difference is more than 4 orders of magnitude, so yes, I expect that $\ce{KHSO4}$ will more strongly favor depronotation than carbonic acid, and hence $\ce{K2SO4}$ is the product I expect.

To get at this quantitatively, consider the sub reaction:

$$\ce{KHCO3 + KHSO4 &lt;=&gt; K2SO4 + H2CO3 }$$ 

Which leads to the expression: 
$$\ce{K_{eq} = \frac{[K2SO4][H2CO3]}{[KHCO3][KHSO4]} }$$ 
which can be rewritten as
$$\ce{K_{eq} = \frac{[H^+][K2SO4][H2CO3]}{[H^+][KHCO3][KHSO4]} }$$ 
Which can be rewritten as 
$$\ce{K_{eq} = \frac{[H2CO3]}{[H^+][KHCO3]}\cdot\frac{[H^+][K2SO4]}{[KHSO4]}}$$ 
which can be written as 
$$\ce{K_{eq} = \frac{$K$_{a2,\ce{KHSO4}}}{$K$_{a1,\ce{H2CO3}}} }$$ 
Taking the log of both sides 
$$\ce{pK_{eq} = p$K$_{a2,\ce{KHSO4}} - p$K$_{a1,\ce{H2CO3}} = } -4.36$$

Which is $$\ce{K_{eq} \approx 23,000}$$ heavily favoring products, and hence, $\ce{K2SO4}$ is the dominant product