Is "Permutation Test" sufficient for a/b testing? I read the "Practical Statistics for Data Scientists" book which is written by Peter Bruce and Andrew Bruce. In this book, the hypothesis testing is only realized with permutation test. If I searched the web, there are formulas, also I did not see too many permutation test blog articles. I want to ask that the permutation test is enough to analyzing a/b experiment data, or do I need different thing ?
Sorry for my grammatical mistakes, my native language is not English.

Consider the two fictitious normal samples below, in which
sample sizes, sample means, and sample variances all differ.
set.seed(2021)
x1 = rnorm( 70, 50,   5)
x2 = rnorm(100, 53.2, 9)

Someone who did not notice the different variances, might
do a pooled 2-sample t test, find no difference in means
at the 5% level, and conclude nothing interesting is going on.
t.test(x1, x2, var.eq=T)$p.val
[1] 0.07286454

summary(x1); length(x1);  sd(x1)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  38.72   45.22   49.70   49.55   52.70   60.60 
[1] 70           # sample size
[1] 5.366852     # sample standard deviation

summary(x2); length(x2);  sd(x2)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  28.89   45.37   51.90   51.87   57.64   77.20 
[1] 100
[1] 9.805152

stripchart(list(x1,x2), ylim=c(.5,2.5), pch="|")


However, a more appropriate Welch 2-sample t test (that does not
assume equal variances) does find a difference in means at the 5% level.
t.test(x1, x2)$p.val
[1] 0.04866419

And (assuming normality) an F-test finds a highly significant difference
between sample variances.
var.test(x1,x2)$p.val
[1] 3.605368e-07

Even the notoriously underpowered Kolmogorov-Smirnov test
finds that the two population distributions are not quite the same.
ks.test(x1,x2)$p.val
[1] 0.04495466

A nonparametric Wilcoxon Rank Sum test finds a significant difference.
However, owing to the different shapes of the samples, this cannot be
considered as a test for difference in medians.
wilcox.test(x1, x2)

        Wilcoxon rank sum test with continuity correction

data:  x1 and x2
W = 2878, p-value = 0.04909
alternative hypothesis: true location shift is not equal to 0

Nor is it clear that the (barely significant) 'location shift' really amounts to stochastic
domination by the second (larger) sample [brown in the empirical
CDF plots below.]
hdr = "ECDFs of Samples 1 [blue] and 2 [brown]"
plot(ecdf(x1), col="blue", main=hdr)
 lines(ecdf(x2), col="brown")


Furthermore, permutation tests using various metrics could be
proposed, which might show a difference in population means---or not.
Depending on what you mean by "better," you might find a standard
or permutation test to support that either A or B is "better"--or that there is not enough "difference" between them to be of practical
importance.
