Why should one not use a 2^p size hash table when using the division method as a hash function?

I don't understand what is meant by:

"m should not be a power of 2, since if m = 2^p, then h(k) is just the p lowest-order bits of k." (pg. 231 of CLRS)

Terms defined:

m: size of hash table
h(k): hash function = k mod m
k: key

I don't understand what the p lowest-order bits of k means. Any clarification would be very helpful.

Refresh your knowledge of binary! The $p$ lowest-order bits of $k$ are the last $p$ bits when $k$ is written out in binary (i.e., the $p$ rightmost bits). For example, if $p=3$ and $k=17$ then $k$ is $10001$ in binary and the three lowest-order bits are $001$.

The point is that, in general, computing $k \bmod m$ is a relatively expensive operation. However, if $m=2^p$ is a power of two, the remainder operation is to just look at the $p$ lowest-order bits. That can be done in a single instruction, by taking the bitwise AND with $2^p-1$.

If it's easier, consider the decimal case. If I ask you what is $8237643\bmod 1034$, your response will be to reach for your calculator; if I ask you what is $8237643\bmod 1000$, you'll tell me that it's $643$ with barely a thought. This is because $1000=10^3$ is a power of ten, whereas $1034$ is not. (There's no immediate equivalent of the bitwise AND trick for decimal.)

StackExchange Computer Science Q#19020, answer score: 17