The standard enthalpy of formation for any substance is the change in enthalpy that accompanies the formation of 1 mole of that substance from its contituent elements in their most stable form.
Take a look at the balanced chemical equation for your reaction
#P_(4(s)) + 10Cl_(2(g)) -> color(red)(4)PCl_(5(g))#
Notice that your reaction produces #color(red)(4)# moles of phosphorus pentachloride. In order for the change in enthalpy, #DeltaH#, to match the standard enthalpy of formation, #DeltaH_f^0#, you need your reaction to produce 1 mole of phosphorus pentachloride.
Divide all the stoichiometric coefficients by 4 to get
#1/4P_(4(s)) + 10/4Cl_(2(g)) -> color(red)(4)/4PCl_(5(g))#
This is equivalent to
#1/4P_(4(s)) + 5/2Cl_(2(g)) -> PCl_(5(g))#
The change in enthalpy for this reaction will be
#DeltaH_"new" = (DeltaH)/4 = "-1776 kJ"/4 = "-444 kJ"#
Since this represents the enthalpy change when 1 mole of phosphorus pentachloride is formed, you can write
#DeltaH_"new" = DeltaH_f^0 = color(green)("-444 kJ/mol")#