The number of P - OH bonds and the oxidation state of phosphorus...

As we can clearly see in the structure,

the number of $P - OH$ bonds is 4.

The oxygen is an electronegative element and will always carry a negative charge. Now, it depends on its surrounding atoms about what charge it carries. In case of a hydroxyl group, the hydrogen carries a +1 charge. So, the opposite charge generated on the oxygen atom along with magnitude will be -1.
In the case of only oxygen atoms, we know that the oxidation number of oxygen is equal to -2. So, the charge generated on the adjacent atom of oxygen will be +2.
Thus, the oxidation state of phosphorus atom will be generated because of:
Oxidation state on phosphorous $= [ 2 \times ( P - OH ) + ( 1 \times = O ) + ( 1 \times - O ) ]$ oxidation states

Thus, the oxidation state of both the phosphorus atoms will be:
${P}_{O . S} = 2 \times ( + 1 ) + 1 \times ( + 2 ) + 1 \times ( + 1 ) = + 5$