You can find the general technique for balancing redox equations in acid solution in this Socratic answer.
We see that #"P"_4# is oxidized to #"H"_3"PO"_4# and #"HOCl"# is reduced to #"Cl"^"-"#.
Step 1: Write the two half-reactions.
#"P"_4 → "H"_3"PO"_4#
#"HOCl" → "Cl"^"-"#
Step 2: Balance all atoms other than #"H"# and #"O"#.
#"P"_4 → "4H"_3"PO"_4#
#"HOCl" → "Cl"^"-"#
Step 3: Balance #"O"#.
#"P"_4 + "16H"_2"O" → "4H"_3"PO"_4#
#"HOCl" → "Cl"^"-" + "H"_2"O"#
Step 4: Balance #"H"#.
#"P"_4 + "16H"_2"O" → "4H"_3"PO"_4+ "20H"^"+" #
#"HOCl" +"H"^"+" → "Cl"^"-" + "H"_2"O"#
Step 5: Balance charge.
#"P"_4 + "16H"_2"O" → "4H"_3"PO"_4+ "20H"^"+" + "20e"^"-"#
#"HOCl" +"H"^"+" + "2e"^"-" → "Cl"^"-" + "H"_2"O"#
Step 6: Equalize electrons transferred.
#1 ×["P"_4 + "16H"_2"O" → "4H"_3"PO"_4+ "20H"^"+" + "20e"^"-"]#
#10 × ["HOCl" +"H"^"+" + "2e"^"-" → "Cl"^"-" + "H"_2"O"]#
Step 7: Add the two half-reactions.
#"P"_4 + stackrelcolor(blue)(6)(color(red)(cancel(color(black)(16))))"H"_2"O" → "4H"_3"PO"_4+ stackrelcolor(blue)(10)(color(red)(cancel(color(black)(20))))"H"^"+" + color(red)(cancel(color(black)(20"e"^"-")))#
#"10HOCl" + color(red)(cancel(color(black)(10"H"^"+"))) + color(red)(cancel(color(black)("20e"^"-"))) → "10Cl"^"-" + color(red)(cancel(color(black)(10"H"_2"O")))#
#stackrel(————————————————————)("P"_4 + 6"H"_2"O" + "10HOCl" → "4H"_3"PO"_4+ 10"H"^"+" + "10Cl"^"-")#
Step 8: Check mass balance.
#mathbf("Atom"color(white)(m)"On the left"color(white)(m)"On the right")#
#color(white)(ml)"P"color(white)(mmmmml)4color(white)(mmmmmmm)4#
#color(white)(ml)"H"color(white)(mmmmll)22color(white)(mmmmmml)22#
#color(white)(ml)"O"color(white)(mmmmll)16color(white)(mmmmmml)16#
#color(white)(ml)"Cl"color(white)(mmmml)10color(white)(mmmmmml)10#
Step 9: Check charge balance.
#mathbf("On the left"color(white)(m)"On the right")#
#color(white)(mmm)0color(white)(mmmll)10 + ("-10") = 0#
∴ The balanced equation is
#"P"_4 + 6"H"_2"O" + "10HOCl" → "4H"_3"PO"_4+ 10"H"^"+" + "10Cl"^"-" #
