The balanced equation is
#"HNO"_3 + "KOH" → "KNO"_3 + "H"_2"O"#
or
#"H"^"+" + "OH"^"-" → "H"_2"O"#
#"Initial moles of H"^+ = "0.040 00" color(red)(cancel(color(black)("L HNO"_3))) × (0.100 color(red)(cancel(color(black)("mol HNO"_3))))/(1 color(red)(cancel(color(black)("L HNO"_3)))) × ("1 mol H"^+)/(1 color(red)(cancel(color(black)("mol HNO"_3)))) = "0.004 00 mol H"^+#
#"Moles of OH"^"-" color(white)(l)"added" = "0.025 00" color(red)(cancel(color(black)("L KOH"))) × (0.100 color(red)(cancel(color(black)("mol KOH"))))/(1 color(red)(cancel(color(black)("L KOH")))) × ("1 mol OH"^"-")/(1 color(red)(cancel(color(black)("mol KOH")))) = "0.002 50 mol OH"^"-"#
You have more moles of #"H"^"+"# than of #"OH"^"-"#, so all of the #"OH"^"-"# will be neutralized.
The moles of #"H"^"+"# remaining are
#"(0.004 00 - 0.002 50) mol" = "0.001 50 mol"#
The total volume is now
#"(40.00 + 25.00) mL = 65.00 mL = 0.065 00 L"#
∴ #["H"^"+"] = "moles"/"litres" = ("0.001 50 mol")/("0.065 00 L") = "0.023 08 mol/L"#
#"pH" = "-log"["H"^"+"] = "-log(0.023 08)" = 1.64#
