Yes, one byte per digit or character.
In hex, X'C1C2C3'.
Depending on the format, I'd tend to agree, ZD the most significant byte, PD the most significant half byte, binary the sing bit then the most significant bit.0011 is a binary 3, the binary string is X'31313334' which I would take as ASCII 1234.
Two bytes of storage can contain 00 thru 99 when usage is ZD, 0 thru 999 when usage is PD, 0 thru 32767 when usage is binary (plus 1 if you ignore the sign).
My recommendation would be to find a different document (at least as far as providing info for an IBM mainframe is concerned). The above is not for the IBM mainframe (EBCDIC character set), but rather how things are stored on UNIX and Win-based systems (ASCII). Care must be taken to make sure the document and the platform are the "same".
For your first question 'ABC' is stored as x'C1C2C3' which in binary is
'1100 0001 1100 0010 1100 0011' (on the mainframe). One byte of alphanumeric data is 2 hexadecimal digits (0-F) or 8 binary digits (0-1). Every 1-byte value is between x'00' and x'FF'.
Back to the info from the document. . .
Let's start with the PICture. For this
+1234 in character format
the PICture in a COBOL program would be PIC S9(4). Internally, it would be stored as x'F1F2F3C4' - note the "C" sign in the low order digit.
The binary value is 1111 0001 1111 0010 1111 0011 1100 0100'
Keep in mind that both the EBCDIC and ASCII character sets are represented by x'00-FF', but the values for the bit patterns is not the same between platforms. This also leads to a difference in collating sequence - the biggest/easiest example is that the numbers 0-9 are lower in the collating sequence than A-Z on ASCII systems, but 0-9 is higher than A-Z on the mainframe.
There are multiple places on the web where you might download both character sets.