drop perpendicular from P to line which is below OP ...let this perpendicular cut this line at Q ....

now we have right angled  triangle POQ in which ,

POQ = @ , PQO = 90^o &  OPQ = 90-@

length of this perpendicular = xsin@

now , we have formula for magnetic field due to current carrieng straight wire (B) = (u_oI/4pia)[sin@₁+sin@₂]

here , a = perpendicular distance

@₁ , @₂ are angles substended with perpendicular line to end points of wire ...

@₁ = @    (from left end it is making @)

@₂ = 90       (right end is at infinity )

now , B = (u_oI/4pi(xsin@) ) .  [sin@+1]

direction of this field is in -ve Z axis ...

in same manner , B due to another straight wire is same ...

total magnetic field = 2B

                             = (u_oI/2pi(xsin@))[1+sin@] (-k)

this is the required ans

@₁ = 90-@ , by mistake i have written @

correct this , then ans will be final ans ..