Use the format() function:

>>> format(14, '#010b')
'0b00001110'

The format() function simply formats the input following the Format Specification mini language. The # makes the format include the 0b prefix, and the 010 size formats the output to fit in 10 characters width, with 0 padding; 2 characters for the 0b prefix, the other 8 for the binary digits.

This is the most compact and direct option.

If you are putting the result in a larger string, use an formatted string literal (3.6+) or use str.format() and put the second argument for the format() function after the colon of the placeholder {:..}:

>>> value = 14
>>> f'The produced output, in binary, is: {value:#010b}'
'The produced output, in binary, is: 0b00001110'
>>> 'The produced output, in binary, is: {:#010b}'.format(value)
'The produced output, in binary, is: 0b00001110'

As it happens, even for just formatting a single value (so without putting the result in a larger string), using a formatted string literal is faster than using format():

>>> import timeit
>>> timeit.timeit("f_(v, '#010b')", "v = 14; f_ = format")  # use a local for performance
0.40298633499332936
>>> timeit.timeit("f'{v:#010b}'", "v = 14")
0.2850222919951193

But I'd use that only if performance in a tight loop matters, as format(...) communicates the intent better.

If you did not want the 0b prefix, simply drop the # and adjust the length of the field:

>>> format(14, '08b')
'00001110'

During phash computation original image is resized. hashSize parameter basically controls height and width of resized image.

Algorithm can be found here. Implementation of the first step (reduce size):

image = image.convert("L").resize((hash_size, hash_size), Image.ANTIALIAS)

See sources of imagehash.phash

Lets see what the line imghash3.append(bin( int(imghash2, 16))[2:].zfill(64)) does.

In [16]: imghash2 = '11b97c7eb158ac'

First of all it converts hexadecimal string into integer

In [17]: int(imghash2, 16)
Out[17]: 4989018956716204

The builtin bin function is applied to convert the integer into a binary string

In [18]: bin( int(imghash2, 16))
Out[18]: '0b10001101110010111110001111110101100010101100010101100'

Drops first two characters using list slice

In [19]: bin( int(imghash2, 16))[2:]
Out[19]: '10001101110010111110001111110101100010101100010101100'

Adds 0 on the left side to make a string of 64 characters total

In [20]: bin( int(imghash2, 16))[2:].zfill(64)
Out[20]: '0000000000010001101110010111110001111110101100010101100010101100'

Strings have a .zfill() method to pad it with zeros:

>>> '100'.zfill(5)
'00100'

For binary numbers however, I'd use string formatting:

>>> '{0:05b}'.format(4)
'00100'

The :05b formatting specification formats the number passed in as binary, with 5 digits, zero padded. See the Python format string syntax. I've used str.format() here, but the built-in format() function can take the same formatting instruction, minus the {0:..} placeholder syntax:

>>> format(4, '05b')
'00100'

if you find that easier.