What will be the filtered output?
First note that:
$$ \cos(2\pi 50 t) \longleftrightarrow 0.5 \delta(f+50) + 0.5\delta(f-50) $$

$$\sin(2\pi 150 t) \longleftrightarrow 0.5 j \delta(f+150) -j 0.5\delta(f-150)$$

Hence the baseband spectrum is: 
$$ X(f) = 0.5 j \delta(f+150) + 0.5 \delta(f+50) + 0.5 \delta(f-50) - 0.5 j\delta(f-150) $$

Then a shift to right by 100 Hz yields (without sampling weight):
$$ X(f-100) = 0.5\left[j \delta(f+50) +\delta(f-50) + \delta(f-150) - j\delta(f-250) \right]$$

Shift left by 100 Hz:
$$ X(f+100) = 0.5\left[ j\delta(f+250) +\delta(f+150) + \delta(f+50) - j\delta(f-50) \right]$$

Shift right by 200 Hz:
$$ X(f-200) = 0.5\left[ j \delta(f-50) +\delta(f-150) + \delta(f-250) - j\delta(f-350) \right]$$

Shift left by 200 Hz:
$$ X(f+200) = 0.5\left[ j\delta(f+350) +\delta(f+250) + \delta(f+150) - j\delta(f+50) \right]$$

The sum of all the shifts for $n=0,\pm 1,\pm2$ :
$$X_s(f) = 0.5\left[
j\delta(f+350) +\delta(f+250) + \delta(f+150) - j\delta(f+50) +
j\delta(f+250) +\delta(f+150) + \delta(f+50) - j\delta(f-50) +
j \delta(f+150) +\delta(f+50) + \delta(f-50) - j\delta(f-150)  +
j \delta(f+50) +\delta(f-50) + \delta(f-150) - j\delta(f-250) +
j \delta(f-50) +\delta(f-150) + \delta(f-250) - j\delta(f-350) \right]
$$

After the 100 Hz lowpass filter only those impulses inside $-100&lt;f&lt;100$ band remain:
$$LPF\{ X_s(f) \} = 0.5\left[ - j\delta(f+50) + \delta(f+50) - j\delta(f-50) +
\delta(f+50) + \delta(f-50) + j \delta(f+50) +\delta(f-50) + j \delta(f-50) 
\right] $$  

Which can be further simplied into:
$$LPF\{ X_s(f) \} =  \delta(f+50) + \delta(f-50)  $$  

Hence we arrive (without sampling weight)
$$ \boxed{ z(t) = 2\cos(2 \pi 50 t)} $$ 

Therefore we can conclude that the output of the LPF will be proportional to $\cos(2\pi 50 t)$.