Maybe this could help

df %>%
  mutate(r = 1:n()) %>%
  unnest(col1) %>%
  table() %>%
  t()

which gives

   col1
r   A B C
  1 1 0 0
  2 1 1 0
  3 0 1 1
  4 0 1 0
  5 1 1 1

Data

df <- tibble(
  col1 = list(
    "A",
    c("A", "B"),
    c("B", "C"),
    "B",
    c("A", "B", "C")
  )
)

If your data is given in the following format

df <- data.frame(
  col1 = c("A", "A,B", "B,C", "B", "A,B,C")
)

then you can try

with(
  df,
  table(rev(stack(setNames(strsplit(col1, ","), seq_along(col1)))))
)

which gives

   values
ind A B C
  1 1 0 0
  2 1 1 0
  3 0 1 1
  4 0 1 0
  5 1 1 1

You could use table() with map_df() from purrr to count the occurrences in each element of a list, and return a data frame. Putting it into a function with some post-processing, and using dplyrs data frame unpacking in mutate(), you could do something like this to stay within a data frame context:

library(tidyverse)

one_hot <- function(x) {
  map_df(x, table) %>% 
    mutate_all(as.integer) %>% 
    mutate_all(replace_na, 0L)
}

df <- data.frame(col1 = c("A", "A,B", "B,C", "B", "A,B,C"))

df %>% 
  mutate(
    one_hot(strsplit(col1, ","))
  )
#>    col1 A B C
#> 1     A 1 0 0
#> 2   A,B 1 1 0
#> 3   B,C 0 1 1
#> 4     B 0 1 0
#> 5 A,B,C 1 1 1

An additional base R solution:

+(
  with(
    df, 
    sapply(
      unique(
        unlist(
          strsplit(
            col1, 
            ","
            )
          )
        ),
      `grepl`, 
      col1
    )
  )
)