No you are not correct. The bit in each byte used for "error detection" is the parity bit. Note that bits are generally counted strangely, using an offset of 1 rather than 0 for the first, leftmost bit - commonly the most significant bit.
So if you'd have a byte array then every least significant bit of every byte is the parity bit. This bit is set in such a way that every byte has an odd number of bits set after the operation. In other words, it is set if the count of the other bits is even and unset if not. Or it is flipped if the amount of all 0 bits or 1 bits is even - which amounts to same thing.
So say you'd have the following single DES key:
pos: 1234 5678 9 .. etc. .... .... .... .... .... .... .... .... .... .... .... ....
bin: 0000_0000 0000_0001 0000_0010 0000_0011 0000_0100 0000_0101 0000_0110 0000_0111
hex: 0 0 0 1 0 2 0 3 0 4 0 5 0 6 0 7
1's: 0 1 1 2 1 2 2 3
then the result is after parity adjustment is this:
pos: 1234 5678 9 .. etc. .... ...P .... ...P .... ...P .... ...P .... ...P .... ...P
bin: 0000_0001 0000_0001 0000_0010 0000_0010 0000_0100 0000_0100 0000_0111 0000_0111
hex: 0 1 0 1 0 2 0 2 0 4 0 4 0 7 0 7
1's: 1 1 1 1 1 1 3 3
So as there are 8 bytes, and one bit of every byte is used for parity, each byte has 7 bits used in the cipher itself. There are 64 / 8 = 8 bytes with 7 bits each, 8 * 7 = 56 bits effective key size.
Parity adjustment and validation depends on the library, but commonly you do need to at least present 8 full bytes / 64 bits.