can i use ttl ic 7400 instead of a 7403 ?
I have consult boths datasheet .only fiff is that 03 use open collector.could not find 7403.

Yes, just omit the pullup resistors that you would have used with the 7403.

Wich one are thoses 3 resistors ? They polarise the onternal fet of the ic ? Should i remove them.all .

R10 only ? Is the pull up or i am wrong .please correct me

In that circuit you can't just drop in a 7400 without adding current limiting resistors on output pins 3 and 6, not that I suggest doing so.

The two NAND gates form an OR gate with R10. This works for open collector outputs but if for example pin 6 is high and pin 3 is low, pin 6 will be shorted to ground. It probably won't blow up but it will be dissipating quite a bit of power in the limiting resistor of the gate with output pin 6.

It'd be better to redesign it and make use of the fourth gate, without having to sum two outputs together as shown above. If you can't do it with four gates, perhaps try a different type of gate instead - you might be able to map it to four NOR gates.

What is U1?
Something funny is going on around D1, there's no path to ground, unless it's through pin 2 of J2?

U1 is an NE567 tone decoder IC. This was a crude 19kHz synchronisation circuit from around 30 years ago!

If you want synchronised 19, 38, 57 and other useful frequencies, it's easiest to use a 4.332 MHz rock (as used in the RDS decoders of some car radios), and a simple set of CMOS dividers. You can also easily do it these days with a 4.332 (or 8.664) MHz crystal and a PIC or AVR. The code is trivially simple.

so removing r10 or leaving everything as it is , should work ? yes that is true it is a pll circuit tone detector .
please confirm Can i use 7400 without the mess of anything or should i remove r10

for the one who said i am a solder jokey and who is the history man ,nothing than words....