The closer an object to the source of light, the larger the shadow it casts.
That's true if we're talking about a point source or at least a compact source of light and "shadow" refers to the "umbra" or area of complete shadowing. But it no longer makes sense in this case where seen from Earth the obscurer (spacecraft) is tiny compared to the "obscuree" (Sun).
In this case we can call the event a transit of the Sun by Parker and can treat it just like a similar transit by Mercury, only smaller.
At the distance of the Earth, There is no umbra, only an antumbra. Parker's umbra only extends about 250 meters behind the spacecraft's 2.3 meter hexagonal Sun shield.
In the case of Mercury, let's do the math.
body radius (km) distance (km) solid angle (sr) relative to Sun
Sun 695,000. 150,000,000. 6.7E-05 -
Mercury 2,440. 92,000,000. 2.1E-09 3.1E-05
Parker 0.0015 150,000,000. 3.1E-22 4.6E-18
So while transit of Mercury will dim the Sun everywhere on Earth almost equally by about 31 parts per million (and would be noticed by good quality photometry from a satellite), Parker would only dim the Sun by five quintillionths, which is far lower than normal fluctuations in the Suns brightness.
At about 195 km/s Parker will transit the Sun's disk in about 7,000 seconds, or 2 hours. Below are examples of how the Sun's brightness fluctuates on this timescale. Certainly a step function of 2E-05 (from Mercury) for hours would be detectable, but one of 3E-12 (from Parker) would be so far in the noise as to be completely undetectable.
However, if you wanted to measure the solar transit of the ISS instead, using a Raspberry Pi or an Arduino and a photodiode from Earth instead, that's certainly doable because of the shortness of the pulse.
below x3: An estimate of the solar background irradiance power spectrum Rabello Soares et al. Astron. Astrophys. 318, 970–974 (1997)